Trigonometric And Hyperbolic Substitutions

When an integrand contains $\sqrt{a^{2} - x^{2}}$ , $\sqrt{a^{2} + x^{2}}$ , or $\sqrt{x^{2} - a^{2}}$ , no ordinary substitution removes the radical. The cure is an inverse substitution: replace $x$ by a trigonometric or hyperbolic function chosen so that a Pythagorean identity collapses the radical into a single term.

Radical	Trigonometric substitution	Hyperbolic substitution	Identity used
$\sqrt{a^{2} - x^{2}}$	$x = a \sin t$ , $- \frac{π}{2} \leq t \leq \frac{π}{2}$	$x = a \tanh t$	$1 - \sin^{2} t = \cos^{2} t$ ; $1 - \tanh^{2} t = {sech}^{2} t$
$\sqrt{a^{2} + x^{2}}$	$x = a \tan θ$ , $- \frac{π}{2} < θ < \frac{π}{2}$	$x = a \sinh t$	$1 + \tan^{2} θ = \sec^{2} θ$ ; $1 + \sinh^{2} t = \cosh^{2} t$
$\sqrt{x^{2} - a^{2}}$	$x = a \sec t$ , $0 \leq t \leq π$ , $t \neq \frac{π}{2}$	$x = \pm a \cosh t$	$\sec^{2} t - 1 = \tan^{2} t$ ; $\cosh^{2} t - 1 = \sinh^{2} t$
$\sqrt{A x^{2} + B x + C}$	Complete the square first, then use one of the rows above

Useful results proved in this section
$\int \frac{d x}{\sqrt{1 + x^{2}}}$	$\ln (x + \sqrt{1 + x^{2}}) + C = arcsinh x + C$
$\int \sqrt{1 + x^{2}} d x$	$\frac{1}{2} x \sqrt{1 + x^{2}} + \frac{1}{2} \ln (x + \sqrt{1 + x^{2}}) + C$
$\int \sqrt{1 - x^{2}} d x$	$\frac{1}{2} \arcsin x + \frac{1}{2} x \sqrt{1 - x^{2}} + C$
$\int \frac{d x}{(a^{2} - x^{2})^{3 / 2}}$	$\frac{x}{a^{2} \sqrt{a^{2} - x^{2}}} + C$

In this section, we learn how to evaluate integrals involving the radicals

\sqrt{a^{2} - x^{2}}, \sqrt{a^{2} + x^{2}}, and \sqrt{x^{2} - a^{2}}

by means of an inverse substitution: we replace $x$ by a function of a new variable $t$ or $θ$ ,

x = ϕ (t) .

We assume throughout that $ϕ$ is one-to-one, so that the substitution is invertible. Choosing $ϕ$ to be a suitable trigonometric or hyperbolic function then eliminates the radical from the integrand.

Integrals Involving the Radical √(a²−x²)

When $\sqrt{1 - x^{2}}$ occurs, to evaluate the integral we look for a substitution

x = ϕ (t) .

Because the expression under the radical, $1 - x^{2}$ , cannot be negative, we must have

- 1 \leq x \leq 1.

A function $ϕ (t)$ whose values vary between $- 1$ and $1$ is the sine (or cosine) function. So we try

x = \sin t (- \frac{π}{2} \leq t \leq \frac{π}{2}) .

We restrict the domain of $ϕ (t) = \sin t$ to $[- \frac{π}{2}, \frac{π}{2}]$ to make the function $ϕ$ one-to-one. Having a one-to-one substitution is important when we want to express the final result in terms of $x$ instead of $t$ . The restriction has a second benefit: on this interval $\cos t \geq 0$ , so $\sqrt{\cos^{2} t} = \cos t$ with no absolute value to worry about.

If $\sqrt{a^{2} - x^{2}} = a \sqrt{1 - {(\frac{x}{a})}^{2}}$ occurs, we can try

x = a \sin t (- \frac{π}{2} \leq t \leq \frac{π}{2})

(or $x = a \cos t$ , $0 \leq t \leq π$ ).

Evaluate $\int \sqrt{1 - x^{2}} d x$ .

Solution

Let $x = \sin θ$ , $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ . Then

d x = \cos θ d θ

and

\begin{aligned} \int\sqrt{1-x^{2}}\,dx &= \int\sqrt{1-\sin^{2}\theta}\ \overbrace{\cos\theta\,d\theta}^{dx}\\ &= \int\sqrt{\cos^{2}\theta}\,\cos\theta\, d\theta &&{\small(\sin^{2}\theta+\cos^{2}\theta=1)}\\ &= \int\cos^{2}\theta\, d\theta &&{\small(\cos\theta\geq0\text{ when }-\tfrac{\pi}{2}\leq\theta\leq\tfrac{\pi}{2})}\\ &= \int\frac{1}{2}(1+\cos2\theta)\,d\theta &&{\small(\cos2\theta=2\cos^{2}\theta-1)}\\ &= \frac{\theta}{2}+\frac{1}{4}\sin2\theta+C. \end{aligned}

Now we need to express the result in terms of $x$ :

x = \sin θ \Rightarrow θ = \arcsin x

and

\begin{aligned} \sin2\theta &= 2\sin\theta\cos\theta\\ &= 2\sin\theta\sqrt{1-\sin^{2}\theta} &&{\small(\cos\theta\geq0\text{ when }-\tfrac{\pi}{2}\leq\theta\leq\tfrac{\pi}{2})}\\ &= 2x\sqrt{1-x^{2}}. \end{aligned}

Therefore

\begin{aligned} \int\sqrt{1-x^{2}}\,dx &= \frac{\theta}{2}+\frac{1}{4}\sin2\theta+C\\ &= \frac{1}{2}\arcsin x+\frac{1}{2}x\sqrt{1-x^{2}}+C. \end{aligned}

Evaluate $\int \frac{\sqrt{4 - x^{2}}}{x^{2}} d x$ .

Solution

The expression under the radical must be nonnegative:

4 - x^{2} \geq 0 \Leftrightarrow - 2 \leq x \leq 2.

Also we must have $x \neq 0$ because $x$ occurs in the denominator. So let $x = 2 \sin θ$ with $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ and $θ \neq 0$ . Then

d x = 2 \cos θ d θ

and

\begin{aligned} \sqrt{4-x^{2}} &= \sqrt{4-4\sin^{2}\theta}\\ &= 2\sqrt{1-\sin^{2}\theta}\\ &= 2\sqrt{\cos^{2}\theta} &&{\small(\sin^{2}\theta+\cos^{2}\theta=1)}\\ &= 2\cos\theta. &&{\small(\cos\theta\geq0\text{ for }-\tfrac{\pi}{2}\leq\theta\leq\tfrac{\pi}{2})} \end{aligned}

Therefore,

\begin{aligned} \int\frac{\sqrt{4-x^{2}}}{x^{2}}\,dx &= \int\frac{2\cos\theta}{4\sin^{2}\theta}\overbrace{2\cos\theta\,d\theta}^{dx}\\ &= \int\frac{\cos^{2}\theta}{\sin^{2}\theta}\,d\theta\\ &= \int\frac{1-\sin^{2}\theta}{\sin^{2}\theta}\,d\theta\\ &= \int\left(\csc^{2}\theta-1\right)\,d\theta\\ &= -\cot\theta-\theta+C. \end{aligned}

Now we have to express the result in terms of $x$ . The second term is easy to convert back: it follows from $\sin θ = x / 2$ that $θ = \arcsin (x / 2)$ . To express $\cot θ$ in terms of $x$ when $\sin θ = x / 2$ , we may use trigonometric identities or geometry.

Using a trigonometric identity. Since $\cos θ \geq 0$ for $- π / 2 \leq θ \leq π / 2$ , $\sin^{2} θ + \cos^{2} θ = 1$ , and $\sin θ = x / 2$ , we have

\cos θ = \sqrt{1 - \sin^{2} θ} = \sqrt{1 - {(\frac{x}{2})}^{2}} .

Therefore,

\begin{aligned} \cot\theta &= \frac{\cos\theta}{\sin\theta}\\ &= \frac{\sqrt{1-\frac{x^{2}}{4}}}{\frac{x}{2}}\\ &= \frac{2}{x}\sqrt{1-\frac{x^{2}}{4}}. \end{aligned}

Using a right-angled triangle. We consider a right triangle in which one of the acute angles is $θ$ . For simplicity, let the hypotenuse be $1$ . Then the side opposite $θ$ must have length $x / 2$ , and by the Pythagorean theorem the side adjacent to $θ$ has length $\sqrt{1 - x^{2} / 4}$ .

A right triangle in which

\sin θ = \frac{x}{2}

, used to read off

\cot θ

Therefore, the cotangent of this angle, which is the ratio of the adjacent side to the opposite side, equals

\begin{aligned} \cot\theta &= \frac{AB}{BC}\\ &= \frac{\sqrt{1-x^{2}/4}}{x/2}\\ &= \frac{2}{x}\sqrt{1-\frac{x^{2}}{4}}. \end{aligned}

Therefore,

\begin{aligned} \int\frac{\sqrt{4-x^{2}}}{x^{2}}\,dx &= -\cot\theta-\theta+C\\ &= -\frac{2}{x}\sqrt{1-\frac{x^{2}}{4}}-\arcsin\left(\frac{x}{2}\right)+C\\ &= -\frac{\sqrt{4-x^{2}}}{x}-\arcsin\left(\frac{x}{2}\right)+C. \end{aligned}

Evaluate $\int x \sqrt{1 - x^{2}} d x$ .

Solution

Method (a). Let $u = 1 - x^{2}$ . Then

d u = - 2 x d x

and

\begin{aligned} \int x\sqrt{1-x^{2}}\,dx &= \int\sqrt{u}\overbrace{\left(\frac{du}{-2}\right)}^{x\,dx}\\ &= -\frac{1}{2}\int\sqrt{u}\,du\\ &= -\frac{1}{2}\times\frac{2}{3}u^{\frac{3}{2}}+C\\ &= -\frac{1}{3}u^{\frac{3}{2}}+C\\ &= -\frac{1}{3}\left(1-x^{2}\right)^{\frac{3}{2}}+C. \end{aligned}

Method (b). Let $x = \sin θ$ , $- \frac{π}{2} \leq θ \leq \frac{π}{2}$ . Then

d x = \cos θ d θ

and

\begin{aligned} \int x\sqrt{1-x^{2}}\,dx &= \int\overbrace{\sin\theta}^{x}\overbrace{\sqrt{1-\sin^{2}\theta}}^{\sqrt{1-x^{2}}}\overbrace{\cos\theta\, d\theta}^{dx}\\ &= \int\sin\theta\sqrt{\cos^{2}\theta}\cos\theta\, d\theta &&{\small(\cos^{2}\theta+\sin^{2}\theta=1)}\\ &= \int\sin\theta\cos^{2}\theta\, d\theta. &&{\small(\cos\theta\geq0\text{ for }-\tfrac{\pi}{2}\leq\theta\leq\tfrac{\pi}{2})} \end{aligned}

Let $u = \cos θ$ , then $d u = - \sin θ d θ$ . Therefore,

\begin{aligned} \int x\sqrt{1-x^{2}}\,dx &= \int\sin\theta\cos^{2}\theta \,d\theta\\ &= \int-u^{2}\,du\\ &= -\frac{1}{3}u^{3}+C\\ &= -\frac{1}{3}\cos^{3}\theta+C &&{\small(u=\cos\theta)}\\ &= -\frac{1}{3}\left(\sqrt{1-\sin^{2}\theta}\right)^{3}+C\\ &= -\frac{1}{3}\left(\sqrt{1-x^{2}}\right)^{3}+C &&{\small(\sin\theta=x)}\\ &= -\frac{1}{3}\left(1-x^{2}\right)^{\frac{3}{2}}+C. \end{aligned}

The moral: when a bare factor of $x$ accompanies the radical, the ordinary substitution $u = a^{2} - x^{2}$ is faster. Reserve the trigonometric substitution for integrands where no such factor is available.

If $\sqrt{a^{2} - x^{2}}$ occurs, because

- 1 < \tanh t < 1 (- \infty < t < + \infty)

we can also try

x = a \tanh t .

\begin{aligned} dx&=a\left(1-\tanh^{2}t\right)\,dt\\ &=a\operatorname{sech}^{2}t\, dt\\ &=\frac{a}{\cosh^{2}t}\,dt. \end{aligned}

However, the trigonometric substitution is often preferred, because the trigonometric functions and the trigonometric identities are more familiar to us.

Using the substitution $x = a \tanh t$ , evaluate $\int \frac{d x}{(a^{2} - x^{2})^{3 / 2}}$ , $| x | < a$ .

Solution

Set $x = a \tanh t$ , so that

d x = a {sech}^{2} t d t .

We simplify $a^{2} - x^{2}$ . Using the identity $1 - \tanh^{2} t = {sech}^{2} t$ ,

a^{2} - x^{2} = a^{2} - a^{2} \tanh^{2} t = a^{2} (1 - \tanh^{2} t) = a^{2} {sech}^{2} t .

Therefore

(a^{2} - x^{2})^{3 / 2} = (a^{2} {sech}^{2} t)^{3 / 2} = a^{3} {sech}^{3} t .

The integral becomes

\int \frac{d x}{(a^{2} - x^{2})^{3 / 2}} = \int \frac{a {sech}^{2} t}{a^{3} {sech}^{3} t} d t = \frac{1}{a^{2}} \int \frac{1}{sech t} d t .

Since $\frac{1}{sech t} = \cosh t$ , this reduces to

\frac{1}{a^{2}} \int \cosh t d t = \frac{\sinh t}{a^{2}} + C .

It remains to express $\sinh t$ in terms of $x$ . From $x = a \tanh t$ we have

\tanh t = \frac{x}{a} .

Recall that $\tanh t = \sinh t / \cosh t$ and $\cosh^{2} t - \sinh^{2} t = 1$ , so $\cosh t = \sqrt{1 + \sinh^{2} t}$ . Write $\sinh t = s$ for brevity. Then

\frac{s}{\sqrt{1 + s^{2}}} = \frac{x}{a} .

Squaring both sides,

\frac{s^{2}}{1 + s^{2}} = \frac{x^{2}}{a^{2}} .

Solving for $s^{2}$ ,

\begin{aligned} a^{2}s^{2} &= x^{2}(1+s^{2}),\\ s^{2}(a^{2}-x^{2}) &= x^{2},\\ s^{2} &= \frac{x^{2}}{a^{2}-x^{2}}. \end{aligned}

Since $x$ and $\sinh t$ have the same sign,

\sinh t = \frac{x}{\sqrt{a^{2} - x^{2}}} .

Substituting back,

\int \frac{d x}{(a^{2} - x^{2})^{3 / 2}} = \frac{1}{a^{2}} \cdot \frac{x}{\sqrt{a^{2} - x^{2}}} + C = \frac{x}{a^{2} \sqrt{a^{2} - x^{2}}} + C .

Notice that we can solve the above example using a trigonometric substitution too. Carrying that out is the first exercise at the end of this section.

Integrals Involving the Radical √(a²+x²)

To evaluate the integrals involving $\sqrt{1 + x^{2}}$ , let us make a substitution

x = ϕ (t) .

Because $\sqrt{1 + x^{2}}$ is defined for every $x$ , the range of $ϕ$ must be all of $ℝ$ . Therefore, we may try the following functions

ϕ (θ) = \tan θ (- \frac{π}{2} < θ < \frac{π}{2})

ϕ (t) = \sinh t (- \infty < t < + \infty),

whose ranges are $ℝ$ .

If $\sqrt{a^{2} + x^{2}} = a \sqrt{1 + (x / a)^{2}}$ occurs, try

x = a \tan θ (- \frac{π}{2} < θ < \frac{π}{2})

x = a \sinh t (- \infty < t < \infty) .

Evaluate $\int \frac{d x}{\sqrt{1 + x^{2}}}$ .

Solution

Method (a). Let $x = \tan θ$ , $- \frac{π}{2} < θ < \frac{π}{2}$ . Then

d x = \sec^{2} θ d θ

and

\begin{aligned} \int\frac{dx}{\sqrt{1+x^{2}}} &= \int\frac{\sec^{2}\theta \,d\theta}{\sqrt{1+\tan^{2}\theta}}\\ &= \int\frac{\sec^{2}\theta}{\sqrt{\sec^{2}\theta}}\,d\theta &&{\small(1+\tan^{2}\theta=\sec^{2}\theta)}\\ &= \int\frac{\sec^{2}\theta}{\sec\theta}\,d\theta &&{\small(\sec\theta>0\text{ when }-\tfrac{\pi}{2}<\theta<\tfrac{\pi}{2})}\\ &= \int\sec\theta \,d\theta\\ &= \ln\left|\sec\theta+\tan\theta\right|+C. \end{aligned}

Now we need to express the result in terms of $x$ . Since $\tan θ = x$ , we just need to express $\sec θ$ in terms of $x$ , and for that we notice that

\sec^{2} θ = 1 + \tan^{2} θ = 1 + x^{2}

and so

\sec θ = \pm \sqrt{1 + x^{2}} .

But which sign should we use? Since we assumed $- π / 2 < θ < π / 2$ , and in this range $\cos θ = 1 / \sec θ > 0$ , we have to choose the plus sign. That is,

\sec θ = \sqrt{1 + x^{2}} .

Therefore,

\begin{aligned} \int\frac{dx}{\sqrt{1+x^{2}}} &= \ln\left|\sec\theta+\tan\theta\right|+C\\ &= \ln\left|\sqrt{1+x^{2}}+x\right|+C\\ &= \ln\left(x+\sqrt{1+x^{2}}\right)+C. \end{aligned}

Here we have dropped the absolute value bars because

\sqrt{1 + x^{2}} > \sqrt{x^{2}} = | x |

and

\sqrt{1 + x^{2}} + x > | x | + x \geq 0.

Method (b). Let $x = \sinh t$ . Then

d x = \cosh t d t

and

\begin{aligned} \int\frac{dx}{\sqrt{1+x^{2}}} &= \int\frac{\overbrace{\cosh t\,dt}^{dx}}{\sqrt{1+\sinh^{2}t}}\\ &= \int\frac{\cosh t}{\sqrt{\cosh^{2}t}}\,dt &&{\small(\cosh^{2}t-\sinh^{2}t=1)}\\ &= \int\frac{\cosh t}{\cosh t}\,dt &&{\small(\text{always }\cosh t>0)}\\ &= \int dt\\ &= t+C\\ &= \operatorname{arcsinh}x+C. &&{\small(x=\sinh t\Leftrightarrow t=\operatorname{arcsinh}x)} \end{aligned}

At first, the results obtained by the first method and by the second method might look different, but we previously showed that

arcsinh x = \ln (x + \sqrt{x^{2} + 1}) .

Evaluate $\int \sqrt{1 + x^{2}} d x$ .

Solution

Method (a). Let $x = \sinh t$ . Then

d x = \cosh t d t

and

\begin{aligned} \int\sqrt{1+x^{2}}\,dx &= \int\sqrt{1+\sinh^{2}t}\,\cosh t\,dt\\ &= \int\sqrt{\cosh^{2}t}\cosh t\,dt &&{\small(\cosh^{2}t-\sinh^{2}t=1)}\\ &= \int\cosh^{2}t\,dt. &&{\small(\cosh t>0)} \end{aligned}

To integrate $\cosh^{2} t$ , we may use the identity

\cosh 2 t = \cosh^{2} t + \sinh^{2} t = 2 \cosh^{2} t - 1

\Rightarrow \cosh^{2} t = \frac{1 + \cosh 2 t}{2} .

Alternatively, we can use the definition of $\cosh t$ , that is

\begin{aligned} \cosh^{2}t &= \left(\frac{e^{t}+e^{-t}}{2}\right)^{2}\\ &= \frac{1}{4}\left(e^{2t}+2e^{t}e^{-t}+e^{-2t}\right)\\ &= \frac{1}{4}\left(e^{2t}+2+e^{-2t}\right). \end{aligned}

Therefore,

\begin{aligned} \int\cosh^{2}t\,dt &= \frac{1}{4}\int\left(e^{2t}+2+e^{-2t}\right)\,dt\\ &= \frac{1}{4}\left(\frac{1}{2}e^{2t}+2t-\frac{1}{2}e^{-2t}\right)+C\\ &= \frac{1}{4}\left(\frac{e^{2t}-e^{-2t}}{2}\right)+\frac{1}{2}t+C\\ &= \frac{1}{4}\sinh(2t)+\frac{1}{2}t+C. \end{aligned}

Because $\sinh 2 t = 2 \sinh t \cosh t$ , we have

\begin{aligned} \int\cosh^{2}t\ dt &= \frac{1}{2}\sinh t\cosh t+\frac{1}{2}t+C\\ &= \frac{1}{2}\sinh t\sqrt{1+\sinh^{2}t}+\frac{1}{2}t+C\\ &= \frac{1}{2}x\sqrt{1+x^{2}}+\frac{1}{2}\operatorname{arcsinh}x+C. &&{\small(x=\sinh t)} \end{aligned}

Therefore

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int\sqrt{1+x^{2}}\,dx=\frac{1}{2}x\sqrt{1+x^{2}}+\frac{1}{2}\operatorname{arcsinh}x+C}\tag{i}

Method (b). Let $x = \tan θ$ and $- \frac{π}{2} < θ < \frac{π}{2}$ . Then

d x = \sec^{2} θ d θ

and

\begin{aligned} \int\sqrt{1+x^{2}}\,dx &= \int\sqrt{1+\tan^{2}\theta}\,\sec^{2}\theta \,d\theta\\ &= \int\sqrt{\sec^{2}\theta}\,\sec^{2}\theta\,d\theta &&{\small(1+\tan^{2}\theta=\sec^{2}\theta)}\\ &= \int\sec^{3}\theta \,d\theta. &&{\small(\sec\theta>0 \text{ for }-\tfrac{\pi}{2}<\theta<\tfrac{\pi}{2})} \end{aligned}

To evaluate $\int \sec^{3} θ d θ$ , we use integration by parts:

\begin{aligned} &u=\sec\theta & &dv=\sec^{2}\theta \,d\theta\\ &du=\sec\theta\tan\theta \,d\theta & &v=\tan\theta. \end{aligned}

Therefore,

\begin{aligned} \int\sec^{3}\theta \,d\theta &= \int\overbrace{\sec\theta}^{u}\ \overbrace{\sec^{2}\theta\,d\theta}^{dv}\\ &= \overbrace{\sec\theta\tan\theta}^{uv}-\int\overbrace{\tan\theta}^{v}\overbrace{\sec\theta\tan\theta \,d\theta}^{du}\\ &= \sec\theta\tan\theta-\int\sec\theta\tan^{2}\theta \,d\theta\\ &= \sec\theta\tan\theta-\int\sec\theta\left(\sec^{2}\theta-1\right)\,d\theta &&{\small(1+\tan^{2}\theta=\sec^{2}\theta)}\\ &= \sec\theta\tan\theta-\int\sec^{3}\theta\, d\theta+\int\sec\theta\, d\theta \end{aligned}

\Rightarrow 2 \int \sec^{3} θ d θ = \sec θ \tan θ + \int \sec θ d θ .

Recall that $\int \sec θ d θ = \ln | \sec θ + \tan θ | + C$ ; therefore

\begin{aligned} \int\sqrt{1+x^{2}}\,dx &= \int\sec^{3}\theta \,d\theta\\ &= \frac{1}{2}\sec\theta\tan\theta+\frac{1}{2}\ln\left|\sec\theta+\tan\theta\right|+C. \end{aligned}

Now we have to express the result in terms of $x$ . Because $\tan θ = x$ , we have

\begin{aligned} \sec\theta &= \sqrt{1+\tan^{2}\theta}\\ &= \sqrt{1+x^{2}} &&{\small(\sec^{2}\theta=1+\tan^{2}\theta\text{ and }\sec\theta>0)} \end{aligned}

\ln | \sec θ + \tan θ | = \ln | \underset{> 0}{\underset{⏟}{x + \sqrt{x^{2} + 1}}} | = \ln (x + \sqrt{x^{2} + 1}) .

We have

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int\sqrt{1+x^{2}}\,dx=\frac{1}{2}x\sqrt{x^{2}+1}+\frac{1}{2}\ln\left(x+\sqrt{x^{2}+1}\right)+C}\tag{ii}

Results (i) and (ii) look quite different, but they are actually the same if we recall that

arcsinh x = \ln (x + \sqrt{x^{2} + 1}) .

Evaluate $\int \frac{d x}{\sqrt{x^{2} + 2 x + 2}}$ .

Solution

In this example, there is a first power of $x$ under the radical, but we can get rid of it by completing the square:

x^{2} + 2 x + 2 = (x + 1)^{2} + 1.

Therefore

\int \frac{d x}{\sqrt{x^{2} + 2 x + 2}} = \int \frac{d u}{\sqrt{u^{2} + 1}} (u = x + 1) .

Method (a). Let $u = \sinh t$ . Then $d u = \cosh t d t$ and

\begin{aligned} \int\frac{du}{\sqrt{u^{2}+1}} &= \int\frac{\cosh t\ dt}{\sqrt{\sinh^{2}t+1}}\\ &= \int\frac{\cosh t}{\cosh t}\,dt &&{\small(\cosh^{2}t-\sinh^{2}t=1,\;\cosh t>0)}\\ &= \int dt\\ &= t+C\\ &= \operatorname{arcsinh}u+C &&{\small(u=\sinh t\Leftrightarrow t=\operatorname{arcsinh}u)}\\ &= \operatorname{arcsinh}(x+1)+C. &&{\small(u=x+1)} \end{aligned}

Method (b). Let $u = \tan θ$ , $- \frac{π}{2} < θ < \frac{π}{2}$ . Then $d u = \sec^{2} θ d θ$ and

\begin{aligned} \int\frac{du}{\sqrt{u^{2}+1}} &= \int\frac{\sec^{2}\theta \,d\theta}{\sqrt{\tan^{2}\theta+1}}\\ &= \int\frac{\sec^{2}\theta}{\sec\theta}\,d\theta &&{\small(\tan^{2}\theta+1=\sec^{2}\theta,\;\sec\theta>0)}\\ &= \int\sec\theta \,d\theta\\ &= \ln\left|\sec\theta+\tan\theta\right|+C. \end{aligned}

If $u = \tan θ$ and $- \frac{π}{2} < θ < \frac{π}{2}$ , then

\sec θ = \sqrt{1 + \tan^{2} θ} = \sqrt{1 + u^{2}} .

Therefore

\begin{aligned} \ln\left|\sec\theta+\tan\theta\right|+C &= \ln\left|u+\sqrt{1+u^{2}}\right|+C\\ &= \ln\left(u+\sqrt{1+u^{2}}\right)+C\\ &= \ln\left(x+1+\sqrt{(x+1)^{2}+1}\right)+C\\ &= \ln\left(x+1+\sqrt{x^{2}+2x+2}\right)+C. \end{aligned}

Why did we drop the absolute value sign above? Because $\sqrt{u^{2} + 1} > | u | \geq - u$ , and therefore $u + \sqrt{u^{2} + 1} > 0$ .

Again, the final results from methods (a) and (b) look different, but since

arcsinh x = \ln (x + \sqrt{x^{2} + 1})

they are mathematically equivalent.

Integrals Involving the Radical √(x²−a²)

When $\sqrt{x^{2} - 1}$ occurs, we must have

x^{2} - 1 \geq 0 \Leftrightarrow x \geq 1 or x \leq - 1.

For $x = ϕ (t)$ to have range $(- \infty, - 1] \cup [1, + \infty)$ , we have two options.

Option (a). We make the substitution

x = \sec t (0 \leq t < \frac{π}{2}) .

(Note that $\sec t$ is not defined for $t = \frac{π}{2}$ .) Then

d x = \sec t \tan t d t

and

\begin{aligned} \sqrt{x^{2}-1} &= \sqrt{\sec^{2}t-1}\\ &= \sqrt{\tan^{2}t} &&{\small(1+\tan^{2}t=\sec^{2}t)}\\ &= \tan t. &&{\small(\tan t\geq0\text{ for }0\leq t<\tfrac{\pi}{2})} \end{aligned}

We can also allow $t$ to take the values $\frac{π}{2} < t \leq π$ . In this case,

\begin{aligned} \sqrt{x^{2}-1} &= \sqrt{\tan^{2}t}\\ &= -\tan t. &&{\small\left(\tan t\leq0\;\text{for}\;\tfrac{\pi}{2}

If $\sqrt{x^{2} - 1}$ appears in the denominator of the integrand, we must exclude the values $t = 0$ and $t = π$ because $\sqrt{x^{2} - 1} = | \tan t | = 0$ for these values. On $[0, π]$ with $t \neq \frac{π}{2}$ , the function $x = \sec t$ increases from $1$ up through all values $\geq 1$ on the first branch, and covers all values $\leq - 1$ on the second branch, so between them the two branches produce exactly the required range.

Option (b). Alternatively, for $x \geq 1$ , we can substitute

x = \cosh t (t \geq 0) .

Then

d x = \sinh t d t

and

\begin{aligned} \sqrt{x^{2}-1} &= \sqrt{\cosh^{2}t-1}\\ &= \sqrt{\sinh^{2}t} &&{\small(\cosh^{2}t-\sinh^{2}t=1)}\\ &= \sinh t. &&{\small(\sinh t\geq0\text{ for }t\geq0)} \end{aligned}

When $x \leq - 1$ , we make the substitution

x = - \cosh t (t \leq 0) .

Then

d x = - \sinh t d t

and

\begin{aligned} \sqrt{x^{2}-1} &= \sqrt{\cosh^{2}t-1}\\ &= \sqrt{\sinh^{2}t}\\ &= |\sinh t|\\ &= -\sinh t. &&{\small(\sinh t\leq0 \text{ for } t\leq0)} \end{aligned}

The graph of $x = \cosh t$ for $t \geq 0$ rises from the point $(0, 1)$ and increases without bound, so it sweeps out exactly the interval $[1, \infty)$ ; its reflection $x = - \cosh t$ for $t \leq 0$ sweeps out $(- \infty, - 1]$ . Together they cover the whole domain of $\sqrt{x^{2} - 1}$ , one branch at a time.

If $\sqrt{x^{2} - a^{2}}$ occurs, try

x = a \sec t 0 \leq t \leq π, t \neq \frac{π}{2}

x=\left\{\begin{aligned} &a\cosh t &&\left(t\geq0\;\text{for}\;x\geq a\right)\\ &-a\cosh t &&(t\leq0\;\text{for}\;x\leq-a). \end{aligned}\right.

Evaluate $\int \frac{d x}{{(4 x^{2} - 9)}^{\frac{3}{2}}}$ .

Solution

Method (a). We can rewrite the denominator of the integrand as

{((\underset{u}{\underset{⏟}{2 x}})^{2} - \underset{a^{2}}{\underset{⏟}{3^{2}}})}^{3 / 2} .

Therefore, we put $u = 2 x = \underset{a}{\underset{⏟}{3}} \sec θ$ , or

x = \frac{3}{2} \sec θ .

Assuming $x > 3 / 2$ restricts our angle to

0 < θ < \frac{π}{2} .

Taking the differential of both sides of $x = \frac{3}{2} \sec θ$ yields

d x = \frac{3}{2} \sec θ \tan θ d θ .

Therefore,

\begin{aligned} \left(4x^{2}-9\right)^{\frac{3}{2}} &= \left(4\times\frac{9}{4}\sec^{2}\theta-9\right)^{\frac{3}{2}}\\ &= 3^{3}\left(\sec^{2}\theta-1\right)^{\frac{3}{2}}\\ &= 27\left(\tan^{2}\theta\right)^{\frac{3}{2}}\\ &= 27\tan^{3}\theta. &&{\small(\tan\theta>0\text{ for }0<\theta<\tfrac{\pi}{2})} \end{aligned}

Therefore,

\begin{aligned} \int\frac{dx}{\left(4x^{2}-9\right)^{\frac{3}{2}}} &= \int\frac{\frac{3}{2}\sec\theta\tan\theta \,d\theta}{27\tan^{3}\theta}\\ &= \frac{1}{18}\int\frac{\sec\theta}{\tan^{2}\theta}\,d\theta\\ &= \frac{1}{18}\int\frac{1}{\cos\theta}\frac{\cos^{2}\theta}{\sin^{2}\theta}\,d\theta\\ &= \frac{1}{18}\int\frac{\cos\theta}{\sin^{2}\theta}\,d\theta. \end{aligned}

Now let $w = \sin θ$ . Then $d w = \cos θ d θ$ and

\begin{aligned} \frac{1}{18}\int\frac{\cos\theta}{\sin^{2}\theta}\,d\theta &= \frac{1}{18}\int w^{-2}\,dw\\ &= -\frac{1}{18}\frac{1}{w}+C\\ &= -\frac{1}{18}\frac{1}{\sin\theta}+C. \end{aligned}

Since $\sec θ = \frac{1}{\cos θ} = \frac{2 x}{3}$ , we have $\cos θ = \frac{3}{2 x}$ , and because $\sin θ > 0$ on $(0, π / 2)$ ,

\sin θ = \sqrt{1 - \frac{9}{4 x^{2}}} = \frac{\sqrt{4 x^{2} - 9}}{2 x}, so \frac{1}{\sin θ} = \frac{2 x}{\sqrt{4 x^{2} - 9}} .

Therefore

\begin{aligned} \int\frac{dx}{\left(4x^{2}-9\right)^{\frac{3}{2}}} &= -\frac{1}{18}\left(\frac{2x}{\sqrt{4x^{2}-9}}\right)+C\\ &= \frac{-x}{9\sqrt{4x^{2}-9}}+C, \end{aligned}

provided that $x > \frac{3}{2}$ .

We can show that the same formula holds if $x < - \frac{3}{2}$ . In fact, if $x < - \frac{3}{2}$ , let $x = \frac{3}{2} \sec θ$ with $\frac{π}{2} < θ < π$ . Then

d x = \frac{3}{2} \sec θ \tan θ d θ

and

\begin{aligned} \left(4x^{2}-9\right)^{\frac{3}{2}} &= 27\left(\tan^{2}\theta\right)^{\frac{3}{2}}\\ &= -27\tan^{3}\theta. &&{\small(\tan\theta<0\;\text{for}\;\tfrac{\pi}{2}<\theta<\pi)} \end{aligned}

Therefore

\begin{aligned} \int\frac{dx}{\left(4x^{2}-9\right)^{\frac{3}{2}}} &= -\frac{1}{18}\int\frac{\cos\theta}{\sin^{2}\theta}\,d\theta\\ &= \frac{1}{18}\frac{1}{\sin\theta}+C. \end{aligned}

If $\sec θ = \frac{1}{\cos θ} = \frac{2 x}{3}$ , then

\begin{aligned} \sin\theta &= \sqrt{1-\cos^{2}\theta} &&{\small(\sin\theta>0\text{ for }\tfrac{\pi}{2}<\theta<\pi)}\\ &= \sqrt{1-\left(\frac{3}{2x}\right)^{2}}\\ &= \sqrt{\frac{4x^{2}-9}{4x^{2}}}\\ &= -\frac{1}{2x}\sqrt{4x^{2}-9}. &&{\small(\sqrt{4x^{2}}=-2x\text{ for }x<-\tfrac{3}{2}<0)} \end{aligned}

Finally

\begin{aligned} \int\frac{dx}{\left(4x^{2}-9\right)^{\frac{3}{2}}} &= \frac{1}{18}\frac{-2x}{\sqrt{4x^{2}-9}}+C\\ &= \frac{-x}{9\sqrt{4x^{2}-9}}+C. \end{aligned}

Method (b). Assuming $x > \frac{3}{2}$ , let $2 x = 3 \cosh t$ , or

x = \frac{3}{2} \cosh t (t > 0) .

Then

d x = \frac{3}{2} \sinh t d t

and

\begin{aligned} \left(4x^{2}-9\right)^{\frac{3}{2}} &= \left(9\cosh^{2}t-9\right)^{\frac{3}{2}}\\ &= 3^{3}\left(\sinh^{2}t\right)^{\frac{3}{2}} &&{\small(\cosh^{2}t-\sinh^{2}t=1)}\\ &= 27\sinh^{3}t. &&{\small(\sinh t>0\text{ for }t>0)} \end{aligned}

Therefore

\begin{aligned} \int\frac{dx}{\left(4x^{2}-9\right)^{\frac{3}{2}}} &= \int\frac{\frac{3}{2}\sinh t\ dt}{27\sinh^{3}t}\\ &= \frac{1}{18}\int\frac{dt}{\sinh^{2}t}\\ &= \frac{1}{18}\int\operatorname{csch}^{2}t\ dt\\ &= -\frac{1}{18}\coth t+C\\ &= -\frac{1}{18}\frac{\cosh t}{\sinh t}+C\\ &= -\frac{1}{18}\frac{\cosh t}{\sqrt{\cosh^{2}t-1}}+C &&{\small(\sinh t>0\text{ for }t>0)}\\ &= -\frac{1}{18}\frac{\frac{2}{3}x}{\sqrt{\frac{4x^{2}}{9}-1}}+C\\ &= \frac{-x}{9\sqrt{4x^{2}-9}}+C, \end{aligned}

as before. When we made the substitution $x = \frac{3}{2} \cosh t$ ( $t \geq 0$ ), we assumed $x > \frac{3}{2}$ . When $x < - \frac{3}{2}$ , we substitute $x = - \frac{3}{2} \cosh t$ . As an exercise, show that this substitution also leads to the same answer.

Exercises

Each of the following can be done with the basic formulas, the substitution rule, the trigonometric integrals of Section 6.1, integration by parts, and the substitutions of this section.

Using the substitution $x = a \sin θ$ , evaluate $\int \frac{d x}{(a^{2} - x^{2})^{3 / 2}}$ , $| x | < a$ .

Solution

Set $x = a \sin θ$ with $θ \in (- \frac{π}{2}, \frac{π}{2})$ , so that

d x = a \cos θ d θ .

We simplify $a^{2} - x^{2}$ . Using the identity $1 - \sin^{2} θ = \cos^{2} θ$ ,

a^{2} - x^{2} = a^{2} - a^{2} \sin^{2} θ = a^{2} (1 - \sin^{2} θ) = a^{2} \cos^{2} θ .

Therefore

(a^{2} - x^{2})^{3 / 2} = (a^{2} \cos^{2} θ)^{3 / 2} = a^{3} \cos^{3} θ .

Because $θ \in (- \frac{π}{2}, \frac{π}{2})$ , we have $\cos θ > 0$ , so no absolute value is needed. The integral becomes

\int \frac{d x}{(a^{2} - x^{2})^{3 / 2}} = \int \frac{a \cos θ d θ}{a^{3} \cos^{3} θ} = \frac{1}{a^{2}} \int \frac{d θ}{\cos^{2} θ} = \frac{1}{a^{2}} \int \sec^{2} θ d θ .

Since $\frac{d}{d θ} \tan θ = \sec^{2} θ$ , we obtain

\frac{1}{a^{2}} \int \sec^{2} θ d θ = \frac{\tan θ}{a^{2}} + C .

It remains to express $\tan θ$ in terms of $x$ . From $x = a \sin θ$ we have

\sin θ = \frac{x}{a} .

Since $\cos θ > 0$ on our interval,

\cos θ = \sqrt{1 - \sin^{2} θ} = \sqrt{1 - \frac{x^{2}}{a^{2}}} = \frac{\sqrt{a^{2} - x^{2}}}{a} .

Therefore

\tan θ = \frac{\sin θ}{\cos θ} = \frac{x / a}{\sqrt{a^{2} - x^{2}} / a} = \frac{x}{\sqrt{a^{2} - x^{2}}} .

Substituting back,

\int \frac{d x}{(a^{2} - x^{2})^{3 / 2}} = \frac{1}{a^{2}} \cdot \frac{x}{\sqrt{a^{2} - x^{2}}} + C = \frac{x}{a^{2} \sqrt{a^{2} - x^{2}}} + C .

Evaluate $\int \frac{d x}{x^{2} \sqrt{1 - x^{2}}}$ .

Answer

- \frac{\sqrt{1 - x^{2}}}{x} + C .

Solution

Method (a). Let $x = \sin θ$ . Then

d x = \cos θ d θ

and

\int \frac{d x}{x^{2} \sqrt{1 - x^{2}}} = \int \frac{\cos θ d θ}{\sin^{2} θ \cos θ} = \int \csc^{2} θ d θ .

Recall that $\frac{d}{d θ} \cot θ = - \csc^{2} θ$ , so

\int \csc^{2} θ d θ = - \cot θ .

Thus

\int \frac{d x}{x^{2} \sqrt{1 - x^{2}}} = - \cot θ + C .

To express $\cot θ$ in terms of $x$ when $\sin θ = x$ , consider a right triangle with one acute angle $θ$ . For the sine of this angle to equal $x$ , let the hypotenuse be $1$ ; then the side opposite the angle must be $x$ , and by the Pythagorean theorem the adjacent side is $\sqrt{1 - x^{2}}$ .

A right triangle in which

\sin θ = x

, used to read off

\cot θ

Therefore the cotangent of this angle, the adjacent side divided by the opposite side, is

\cot θ = \frac{A C}{B C} = \frac{\sqrt{1 - x^{2}}}{x},

and therefore

\int \frac{d x}{x^{2} \sqrt{1 - x^{2}}} = - \frac{\sqrt{1 - x^{2}}}{x} + C .

Method (b). Let $x = \tanh t$ . Then

d x = {sech}^{2} t d t

and

\int \frac{d x}{x^{2} \sqrt{1 - x^{2}}} = \int \frac{{sech}^{2} t d t}{\tanh^{2} t \sqrt{1 - \tanh^{2} t}} .

Since $1 - \tanh^{2} t = {sech}^{2} t$ and $sech t > 0$ ,

\int \frac{{sech}^{2} t d t}{\tanh^{2} t \cdot sech t} = \int \frac{sech t}{\tanh^{2} t} d t .

To simplify, let us write everything in terms of $\sinh t$ and $\cosh t$ :

\int \frac{\frac{1}{\cosh t}}{\frac{\sinh^{2} t}{\cosh^{2} t}} d t = \int \frac{\cosh t}{\sinh^{2} t} d t .

Let $u = \sinh t$ , so $d u = \cosh t d t$ . Then

\int \frac{d u}{u^{2}} = - \frac{1}{u} + C = - \frac{1}{\sinh t} + C .

It remains to express $\sinh t$ in terms of $x$ . Since $\tanh t = x$ , we have

\tanh t = \frac{\sinh t}{\cosh t} = x .

Using $\cosh^{2} t - \sinh^{2} t = 1$ gives $\cosh t = \sqrt{1 + \sinh^{2} t}$ , so

\frac{\sinh t}{\sqrt{1 + \sinh^{2} t}} = x

⟹ \sinh^{2} t = x^{2} (1 + \sinh^{2} t)

⟹ (1 - x^{2}) \sinh^{2} t = x^{2},

and thus

\sinh t = \frac{x}{\sqrt{1 - x^{2}}} .

Therefore

\int \frac{d x}{x^{2} \sqrt{1 - x^{2}}} = - \frac{1}{\sinh t} + C = - \frac{\sqrt{1 - x^{2}}}{x} + C .

Evaluate $\int \frac{\sqrt{4 - x^{2}}}{x^{2}} d x$ using the substitution $x = 2 \tanh t$ .

Answer

- \frac{\sqrt{4 - x^{2}}}{x} - \arcsin (\frac{x}{2}) + C

or, equivalently,

- \frac{\sqrt{4 - x^{2}}}{x} - \arctan (\frac{x}{\sqrt{4 - x^{2}}}) + C .

Solution

The expression under the radical must be nonnegative:

4 - x^{2} \geq 0 \Leftrightarrow - 2 \leq x \leq 2.

Also we must have $x \neq 0$ because $x$ occurs in the denominator. The substitution $x = 2 \tanh t$ covers the open interval $- 2 < x < 2$ , and $x \neq 0$ forces $t \neq 0$ . Then

d x = 2 {sech}^{2} t d t

and

\begin{aligned} \sqrt{4-x^{2}} &= \sqrt{4-4\tanh^{2}t}\\ &= 2\sqrt{1-\tanh^{2}t}\\ &= 2\sqrt{\operatorname{sech}^{2}t} &&{\small(1-\tanh^{2}t=\operatorname{sech}^{2}t)}\\ &= 2\operatorname{sech}t. &&{\small(\operatorname{sech}t>0\text{ for all }t)} \end{aligned}

Notice that $sech t$ is positive everywhere, so unlike the sine substitution there is no sign condition to check. Therefore,

\begin{aligned} \int\frac{\sqrt{4-x^{2}}}{x^{2}}\,dx &= \int\frac{2\operatorname{sech}t}{4\tanh^{2}t}\overbrace{2\operatorname{sech}^{2}t\,dt}^{dx}\\ &= \int\frac{\operatorname{sech}^{3}t}{\tanh^{2}t}\,dt\\ &= \int\frac{dt}{\cosh t\,\sinh^{2}t}. \end{aligned}

To integrate this, multiply numerator and denominator by $\cosh t$ and use $\cosh^{2} t = 1 + \sinh^{2} t$ :

\int \frac{d t}{\cosh t \sinh^{2} t} = \int \frac{\cosh t}{(1 + \sinh^{2} t) \sinh^{2} t} d t .

Now substitute $w = \sinh t$ , so that $d w = \cosh t d t$ :

\begin{aligned} \int&\frac{\cosh t}{(1+\sinh^{2}t)\sinh^{2}t}\,dt \\ &= \int\frac{dw}{(1+w^{2})w^{2}}\\ &= \int\left(\frac{1}{w^{2}}-\frac{1}{1+w^{2}}\right)\,dw &&{\small\left(\tfrac{1}{w^{2}(1+w^{2})}=\tfrac{1}{w^{2}}-\tfrac{1}{1+w^{2}}\right)}\\ &= -\frac{1}{w}-\arctan w+C\\ &= -\frac{1}{\sinh t}-\arctan(\sinh t)+C. &&{\small(w=\sinh t)} \end{aligned}

The algebraic identity used on the third line is verified by putting the right side over a common denominator: $\frac{1}{w^{2}} - \frac{1}{1 + w^{2}} = \frac{(1 + w^{2}) - w^{2}}{w^{2} (1 + w^{2})} = \frac{1}{w^{2} (1 + w^{2})}$ .

Now we express the result in terms of $x$ . From $\tanh t = x / 2$ and $\cosh t > 0$ we have

\cosh t = \frac{1}{\sqrt{1 - \tanh^{2} t}} = \frac{1}{\sqrt{1 - \frac{x^{2}}{4}}} = \frac{2}{\sqrt{4 - x^{2}}},

and therefore

\sinh t = \tanh t \cosh t = \frac{x}{2} \cdot \frac{2}{\sqrt{4 - x^{2}}} = \frac{x}{\sqrt{4 - x^{2}}} .

Substituting back gives

\begin{aligned} \int\frac{\sqrt{4-x^{2}}}{x^{2}}\,dx &= -\frac{1}{\sinh t}-\arctan(\sinh t)+C\\ &= -\frac{\sqrt{4-x^{2}}}{x}-\arctan\left(\frac{x}{\sqrt{4-x^{2}}}\right)+C. \end{aligned}

This agrees with the answer obtained by the sine substitution. Indeed, if $θ = \arcsin (x / 2)$ then $\tan θ = \frac{x / 2}{\sqrt{1 - x^{2} / 4}} = \frac{x}{\sqrt{4 - x^{2}}}$ , so

\arctan (\frac{x}{\sqrt{4 - x^{2}}}) = \arcsin (\frac{x}{2}),

and the result can be written in the same form as before:

\int \frac{\sqrt{4 - x^{2}}}{x^{2}} d x = - \frac{\sqrt{4 - x^{2}}}{x} - \arcsin (\frac{x}{2}) + C .

Let $a > 0$ . Evaluate $\int \frac{d x}{x^{2} \sqrt{x^{2} + a^{2}}}$ .

Solution

Method 1: trigonometric substitution. Put $x = a \tan θ$ with $- \frac{π}{2} < θ < \frac{π}{2}$ . On this interval $\sec θ > 0$ , so

d x = a \sec^{2} θ d θ,

\sqrt{x^{2} + a^{2}} = a \sqrt{\tan^{2} θ + 1} = a \sec θ,

x^{2} = a^{2} \tan^{2} θ .

Substitute everything in:

\int \frac{a \sec^{2} θ d θ}{a^{2} \tan^{2} θ \cdot a \sec θ} = \frac{1}{a^{2}} \int \frac{\sec θ}{\tan^{2} θ} d θ .

Rewrite the integrand through sines and cosines:

\frac{\sec θ}{\tan^{2} θ} = \frac{1 / \cos θ}{\sin^{2} θ / \cos^{2} θ} = \frac{\cos θ}{\sin^{2} θ} .

With $u = \sin θ$ , $d u = \cos θ d θ$ ,

\frac{1}{a^{2}} \int \frac{d u}{u^{2}} = - \frac{1}{a^{2} u} + C = - \frac{1}{a^{2} \sin θ} + C .

To return to $x$ , read off a right triangle with $\tan θ = \frac{x}{a}$ : the opposite side is $x$ , the adjacent side is $a$ , and the hypotenuse is $\sqrt{x^{2} + a^{2}}$ .

A right triangle in which

\tan θ = \frac{x}{a}

, used to read off

\sin θ

Hence

\sin θ = \frac{B C}{A C} = \frac{x}{\sqrt{x^{2} + a^{2}}},

and therefore

\int \frac{d x}{x^{2} \sqrt{x^{2} + a^{2}}} = - \frac{\sqrt{x^{2} + a^{2}}}{a^{2} x} + C .

Method 2: hyperbolic substitution. Put $x = a \sinh t$ . Since $\cosh t > 0$ for all $t$ ,

d x = a \cosh t d t,

\sqrt{x^{2} + a^{2}} = a \sqrt{\sinh^{2} t + 1} = a \cosh t,

x^{2} = a^{2} \sinh^{2} t .

The integral collapses nicely:

\int \frac{a \cosh t d t}{a^{2} \sinh^{2} t \cdot a \cosh t} = \frac{1}{a^{2}} \int \frac{d t}{\sinh^{2} t} = \frac{1}{a^{2}} \int {csch}^{2} t d t .

Because $\frac{d}{d t} \coth t = - {csch}^{2} t$ ,

\frac{1}{a^{2}} \int {csch}^{2} t d t = - \frac{\coth t}{a^{2}} .

Now undo the substitution. From $\sinh t = \frac{x}{a}$ ,

\cosh t = \sqrt{1 + \sinh^{2} t} = \frac{\sqrt{x^{2} + a^{2}}}{a},

\coth t = \frac{\cosh t}{\sinh t} = \frac{\sqrt{x^{2} + a^{2}}}{x} .

This gives the same antiderivative as before:

\int \frac{d x}{x^{2} \sqrt{x^{2} + a^{2}}} = - \frac{\sqrt{x^{2} + a^{2}}}{a^{2} x} + C .

Check. Differentiate the result. With $g (x) = \sqrt{x^{2} + a^{2}}$ and g'(x)=\dfrac{x}{\sqrt{x^{2}+a^{2}}},

\frac{d}{dx}\!\left(\frac{g}{x}\right) = \frac{g'x-g}{x^{2}} = \frac{-a^{2}}{x^{2}\sqrt{x^{2}+a^{2}}},

\frac{d}{d x} (- \frac{\sqrt{x^{2} + a^{2}}}{a^{2} x}) = \frac{1}{x^{2} \sqrt{x^{2} + a^{2}}},

which recovers the integrand.

Evaluate $\int \sqrt{x^{2} - 1} d x$ , $| x | > 1$ .

Solution

Method 1: trigonometric substitution. Take the branch $x \geq 1$ and put $x = \sec θ$ with $0 \leq θ < \frac{π}{2}$ , where $\tan θ \geq 0$ . Then

d x = \sec θ \tan θ d θ,

\sqrt{x^{2} - 1} = \sqrt{\sec^{2} θ - 1} = \tan θ .

The integral becomes

\int \tan θ \cdot \sec θ \tan θ d θ = \int \sec θ \tan^{2} θ d θ .

Use $\tan^{2} θ = \sec^{2} θ - 1$ :

\int \sec^{3} θ d θ - \int \sec θ d θ .

The base integral is standard,

\int \sec θ d θ = \int \frac{\sec θ (\sec θ + \tan θ)}{\sec θ + \tan θ} d θ = \ln | \sec θ + \tan θ |,

since the numerator is the derivative of the denominator. For the cubic power, integrate by parts with $u = \sec θ$ , $d v = \sec^{2} θ d θ$ , so $v = \tan θ$ :

\int \sec^{3} θ d θ = \sec θ \tan θ - \int \sec θ \tan^{2} θ d θ .

Replace $\tan^{2} θ = \sec^{2} θ - 1$ on the right and collect the $\int \sec^{3} θ$ terms:

2 \int \sec^{3} θ d θ = \sec θ \tan θ + \ln | \sec θ + \tan θ | .

Hence

\int \sec^{3} θ d θ = \frac{\sec θ \tan θ}{2} + \frac{1}{2} \ln | \sec θ + \tan θ | .

Subtracting $\int \sec θ d θ$ ,

I = \frac{\sec θ \tan θ}{2} - \frac{1}{2} \ln | \sec θ + \tan θ | .

Return to $x$ through $\sec θ = x$ , $\tan θ = \sqrt{x^{2} - 1}$ :

\int \sqrt{x^{2} - 1} d x = \frac{x \sqrt{x^{2} - 1}}{2} - \frac{1}{2} \ln | x + \sqrt{x^{2} - 1} | + C .

Method 2: hyperbolic substitution. Take $x \geq 1$ and put $x = \cosh t$ with $t \geq 0$ , so that $\sinh t \geq 0$ . Then

d x = \sinh t d t,

\sqrt{x^{2} - 1} = \sqrt{\cosh^{2} t - 1} = \sinh t .

The integral becomes

\int \sinh t \cdot \sinh t d t = \int \sinh^{2} t d t .

Apply $\sinh^{2} t = \frac{1}{2} (\cosh 2 t - 1)$ :

\int \sinh^{2} t d t = \frac{1}{2} \int (\cosh 2 t - 1) d t = \frac{\sinh 2 t}{4} - \frac{t}{2} .

Now undo the substitution. With $\cosh t = x$ and $\sinh t = \sqrt{x^{2} - 1}$ ,

\sinh 2 t = 2 \sinh t \cosh t = 2 x \sqrt{x^{2} - 1},

and $t = arccosh x = \ln (x + \sqrt{x^{2} - 1})$ . Therefore

\frac{\sinh 2 t}{4} - \frac{t}{2} = \frac{x \sqrt{x^{2} - 1}}{2} - \frac{1}{2} \ln (x + \sqrt{x^{2} - 1}),

which agrees with Method 1.

Method 3: integration by parts. No substitution is required. Take $u = \sqrt{x^{2} - 1}$ and $d v = d x$ , so that

d u = \frac{x}{\sqrt{x^{2} - 1}} d x, v = x .

Then

\int \sqrt{x^{2} - 1} d x = x \sqrt{x^{2} - 1} - \int \frac{x^{2}}{\sqrt{x^{2} - 1}} d x .

Split the numerator as $x^{2} = (x^{2} - 1) + 1$ :

\int \frac{x^{2}}{\sqrt{x^{2} - 1}} d x = \int \sqrt{x^{2} - 1} d x + \int \frac{d x}{\sqrt{x^{2} - 1}} .

The last integral is standard,

\int \frac{d x}{\sqrt{x^{2} - 1}} = \ln | x + \sqrt{x^{2} - 1} | .

Writing $I = \int \sqrt{x^{2} - 1} d x$ , the previous lines say

I = x \sqrt{x^{2} - 1} - I - \ln | x + \sqrt{x^{2} - 1} | .

Solve for $I$ :

2 I = x \sqrt{x^{2} - 1} - \ln | x + \sqrt{x^{2} - 1} |,

\int \sqrt{x^{2} - 1} d x = \frac{x \sqrt{x^{2} - 1}}{2} - \frac{1}{2} \ln | x + \sqrt{x^{2} - 1} | + C,

the same result a third time.

Check. Differentiate the answer. With $v = \sqrt{x^{2} - 1}$ and v'=\dfrac{x}{\sqrt{x^{2}-1}},

\frac{d}{dx}\!\left(\frac{xv}{2}\right)=\frac{v}{2}+\frac{x v'}{2}=\frac{\sqrt{x^{2}-1}}{2}+\frac{x^{2}}{2\sqrt{x^{2}-1}}=\frac{2x^{2}-1}{2\sqrt{x^{2}-1}}.

Together with $\frac{d}{d x} \ln | x + \sqrt{x^{2} - 1} | = \frac{1}{\sqrt{x^{2} - 1}}$ ,

\frac{d}{d x} (\frac{x \sqrt{x^{2} - 1}}{2} - \frac{1}{2} \ln | x + \sqrt{x^{2} - 1} |) = \frac{2 x^{2} - 1 - 1}{2 \sqrt{x^{2} - 1}} = \frac{x^{2} - 1}{\sqrt{x^{2} - 1}} = \sqrt{x^{2} - 1},

which is the original integrand.

Evaluate $\int \frac{d x}{\sqrt{9 - x^{2}}}$ .

Solution

Let $x = 3 \sin θ$ with $- \frac{π}{2} < θ < \frac{π}{2}$ , so $d x = 3 \cos θ d θ$ and $\sqrt{9 - x^{2}} = 3 \cos θ$ (positive on this interval). Then

\int \frac{d x}{\sqrt{9 - x^{2}}} = \int \frac{3 \cos θ}{3 \cos θ} d θ = θ + C = \arcsin \frac{x}{3} + C .

Evaluate $\int \sqrt{9 - x^{2}} d x$ .

Solution

Let $x = 3 \sin θ$ , $d x = 3 \cos θ d θ$ , $\sqrt{9 - x^{2}} = 3 \cos θ$ . Then

\begin{aligned} \int\sqrt{9-x^{2}}\,dx &= \int 9\cos^{2}\theta\,d\theta\\ &= \frac{9}{2}\int(1+\cos 2\theta)\,d\theta\\ &= \frac{9}{2}\theta+\frac{9}{4}\sin 2\theta+C. \end{aligned}

Since $\sin θ = \frac{x}{3}$ and $\cos θ = \frac{\sqrt{9 - x^{2}}}{3}$ ,

\sin 2 θ = 2 \sin θ \cos θ = \frac{2 x \sqrt{9 - x^{2}}}{9} .

Therefore

\int \sqrt{9 - x^{2}} d x = \frac{9}{2} \arcsin \frac{x}{3} + \frac{x \sqrt{9 - x^{2}}}{2} + C .

Evaluate $\int \frac{x^{2}}{\sqrt{1 - x^{2}}} d x$ .

Solution

Let $x = \sin θ$ , so $d x = \cos θ d θ$ and $\sqrt{1 - x^{2}} = \cos θ$ . Then

\begin{aligned} \int\frac{x^{2}}{\sqrt{1-x^{2}}}\,dx &= \int\sin^{2}\theta\,d\theta\\ &= \frac{\theta}{2}-\frac{\sin 2\theta}{4}+C\\ &= \frac{\theta}{2}-\frac{\sin\theta\cos\theta}{2}+C. \end{aligned}

Since $\sin θ = x$ and $\cos θ = \sqrt{1 - x^{2}}$ ,

\int \frac{x^{2}}{\sqrt{1 - x^{2}}} d x = \frac{1}{2} \arcsin x - \frac{x \sqrt{1 - x^{2}}}{2} + C .

Evaluate $\int x^{2} \sqrt{4 - x^{2}} d x$ .

Solution

Let $x = 2 \sin θ$ , so $d x = 2 \cos θ d θ$ and $\sqrt{4 - x^{2}} = 2 \cos θ$ . Then

\begin{aligned} \int x^{2}\sqrt{4-x^{2}}\,dx &= \int 4\sin^{2}\theta\cdot2\cos\theta\cdot2\cos\theta\,d\theta\\ &= 16\int\sin^{2}\theta\cos^{2}\theta\,d\theta. \end{aligned}

From Section 6.1, $\int \sin^{2} θ \cos^{2} θ d θ = \frac{θ}{8} - \frac{\sin 4 θ}{32}$ , so the integral equals $2 θ - \frac{\sin 4 θ}{2}$ .

Now convert back. With $\sin θ = \frac{x}{2}$ and $\cos θ = \frac{\sqrt{4 - x^{2}}}{2}$ ,

\sin 2 θ = 2 \sin θ \cos θ = \frac{x \sqrt{4 - x^{2}}}{2}, \cos 2 θ = 1 - 2 \sin^{2} θ = 1 - \frac{x^{2}}{2},

\sin 4 θ = 2 \sin 2 θ \cos 2 θ = x \sqrt{4 - x^{2}} (1 - \frac{x^{2}}{2}) = \frac{x (2 - x^{2}) \sqrt{4 - x^{2}}}{2} .

Therefore

\int x^{2} \sqrt{4 - x^{2}} d x = 2 \arcsin \frac{x}{2} - \frac{x (2 - x^{2}) \sqrt{4 - x^{2}}}{4} + C .

Evaluate $\int \frac{d x}{x \sqrt{1 - x^{2}}}$ .

Solution

Let $x = \sin θ$ , so $d x = \cos θ d θ$ and $\sqrt{1 - x^{2}} = \cos θ$ . Then

\int \frac{d x}{x \sqrt{1 - x^{2}}} = \int \frac{\cos θ d θ}{\sin θ \cos θ} = \int \csc θ d θ = - \ln | \csc θ + \cot θ | + C .

Now $\csc θ = \frac{1}{x}$ and $\cot θ = \frac{\sqrt{1 - x^{2}}}{x}$ , so

\int \frac{d x}{x \sqrt{1 - x^{2}}} = - \ln | \frac{1 + \sqrt{1 - x^{2}}}{x} | + C .

Evaluate $\int \frac{d x}{x^{2} \sqrt{4 - x^{2}}}$ .

Solution

Let $x = 2 \sin θ$ , so $d x = 2 \cos θ d θ$ and $\sqrt{4 - x^{2}} = 2 \cos θ$ . Then

\begin{aligned} \int\frac{dx}{x^{2}\sqrt{4-x^{2}}} &= \int\frac{2\cos\theta\,d\theta}{4\sin^{2}\theta\cdot2\cos\theta}\\ &= \frac{1}{4}\int\csc^{2}\theta\,d\theta\\ &= -\frac{\cot\theta}{4}+C. \end{aligned}

With $\sin θ = \frac{x}{2}$ and $\cos θ = \frac{\sqrt{4 - x^{2}}}{2}$ we get $\cot θ = \frac{\sqrt{4 - x^{2}}}{x}$ , so

\int \frac{d x}{x^{2} \sqrt{4 - x^{2}}} = - \frac{\sqrt{4 - x^{2}}}{4 x} + C .

Evaluate $\int \sqrt{1 - 4 x^{2}} d x$ .

Solution

First make the linear substitution $u = 2 x$ , so $d x = \frac{1}{2} d u$ :

\int \sqrt{1 - 4 x^{2}} d x = \frac{1}{2} \int \sqrt{1 - u^{2}} d u .

We showed in the first example of this section that $\int \sqrt{1 - u^{2}} d u = \frac{1}{2} \arcsin u + \frac{1}{2} u \sqrt{1 - u^{2}} + C$ . Therefore

\begin{aligned} \int\sqrt{1-4x^{2}}\,dx &= \frac{1}{4}\arcsin(2x)+\frac{1}{4}(2x)\sqrt{1-4x^{2}}+C\\ &= \frac{1}{4}\arcsin(2x)+\frac{x}{2}\sqrt{1-4x^{2}}+C. \end{aligned}

Evaluate $\int \frac{d x}{(x^{2} + 4)^{3 / 2}}$ .

Solution

Let $x = 2 \tan θ$ with $- \frac{π}{2} < θ < \frac{π}{2}$ , so $d x = 2 \sec^{2} θ d θ$ and $x^{2} + 4 = 4 \sec^{2} θ$ , hence $(x^{2} + 4)^{3 / 2} = 8 \sec^{3} θ$ (using $\sec θ > 0$ ). Then

\begin{aligned} \int\frac{dx}{(x^{2}+4)^{3/2}} &= \int\frac{2\sec^{2}\theta\,d\theta}{8\sec^{3}\theta}\\ &= \frac{1}{4}\int\cos\theta\,d\theta\\ &= \frac{\sin\theta}{4}+C. \end{aligned}

From $\tan θ = \frac{x}{2}$ we read $\sin θ = \frac{x}{\sqrt{x^{2} + 4}}$ , so

\int \frac{d x}{(x^{2} + 4)^{3 / 2}} = \frac{x}{4 \sqrt{x^{2} + 4}} + C .

Evaluate $\int \frac{d x}{x \sqrt{x^{2} + 4}}$ .

Solution

Let $x = 2 \tan θ$ , so $d x = 2 \sec^{2} θ d θ$ and $\sqrt{x^{2} + 4} = 2 \sec θ$ . Then

\begin{aligned} \int\frac{dx}{x\sqrt{x^{2}+4}} &= \int\frac{2\sec^{2}\theta\,d\theta}{2\tan\theta\cdot2\sec\theta}\\ &= \frac{1}{2}\int\frac{\sec\theta}{\tan\theta}\,d\theta\\ &= \frac{1}{2}\int\csc\theta\,d\theta\\ &= -\frac{1}{2}\ln|\csc\theta+\cot\theta|+C. \end{aligned}

Since $\csc θ = \frac{\sqrt{x^{2} + 4}}{x}$ and $\cot θ = \frac{2}{x}$ ,

\int \frac{d x}{x \sqrt{x^{2} + 4}} = - \frac{1}{2} \ln | \frac{2 + \sqrt{x^{2} + 4}}{x} | + C .

Evaluate $\int \sqrt{4 + x^{2}} d x$ .

Solution

Let $x = 2 \sinh t$ , so $d x = 2 \cosh t d t$ and $\sqrt{4 + x^{2}} = 2 \cosh t$ . Then

\begin{aligned} \int\sqrt{4+x^{2}}\,dx &= \int 4\cosh^{2}t\,dt\\ &= 2\int(1+\cosh 2t)\,dt\\ &= 2t+\sinh 2t+C\\ &= 2t+2\sinh t\cosh t+C. \end{aligned}

Now $\sinh t = \frac{x}{2}$ , $\cosh t = \frac{\sqrt{4 + x^{2}}}{2}$ , and $t = arcsinh \frac{x}{2} = \ln (\frac{x + \sqrt{4 + x^{2}}}{2})$ . Therefore

\int \sqrt{4 + x^{2}} d x = \frac{x \sqrt{4 + x^{2}}}{2} + 2 \ln (x + \sqrt{4 + x^{2}}) + C,

where the constant $- 2 \ln 2$ has been absorbed into $C$ .

Evaluate $\int \frac{x^{2}}{(x^{2} + 1)^{2}} d x$ .

Solution

Let $x = \tan θ$ , so $d x = \sec^{2} θ d θ$ and $(x^{2} + 1)^{2} = \sec^{4} θ$ . Then

\begin{aligned} \int\frac{x^{2}}{(x^{2}+1)^{2}}\,dx &= \int\frac{\tan^{2}\theta\sec^{2}\theta}{\sec^{4}\theta}\,d\theta\\ &= \int\frac{\tan^{2}\theta}{\sec^{2}\theta}\,d\theta\\ &= \int\sin^{2}\theta\,d\theta\\ &= \frac{\theta}{2}-\frac{\sin\theta\cos\theta}{2}+C. \end{aligned}

From $\tan θ = x$ we get $\sin θ = \frac{x}{\sqrt{1 + x^{2}}}$ and $\cos θ = \frac{1}{\sqrt{1 + x^{2}}}$ , so $\sin θ \cos θ = \frac{x}{1 + x^{2}}$ . Therefore

\int \frac{x^{2}}{(x^{2} + 1)^{2}} d x = \frac{1}{2} \arctan x - \frac{x}{2 (1 + x^{2})} + C .

Evaluate $\int \frac{d x}{(9 + x^{2})^{2}}$ .

Solution

Let $x = 3 \tan θ$ , so $d x = 3 \sec^{2} θ d θ$ and $(9 + x^{2})^{2} = 81 \sec^{4} θ$ . Then

\begin{aligned} \int\frac{dx}{(9+x^{2})^{2}} &= \int\frac{3\sec^{2}\theta\,d\theta}{81\sec^{4}\theta}\\ &= \frac{1}{27}\int\cos^{2}\theta\,d\theta\\ &= \frac{1}{27}\left(\frac{\theta}{2}+\frac{\sin\theta\cos\theta}{2}\right)+C. \end{aligned}

Since $\sin θ = \frac{x}{\sqrt{x^{2} + 9}}$ and $\cos θ = \frac{3}{\sqrt{x^{2} + 9}}$ , we have $\sin θ \cos θ = \frac{3 x}{x^{2} + 9}$ , so

\int \frac{d x}{(9 + x^{2})^{2}} = \frac{1}{54} \arctan \frac{x}{3} + \frac{x}{18 (x^{2} + 9)} + C .

Evaluate $\int \frac{x^{3}}{\sqrt{x^{2} + 9}} d x$ in two ways: by an ordinary substitution, and by a trigonometric substitution.

Solution

Ordinary substitution. Let $u = x^{2} + 9$ , so $x^{2} = u - 9$ and $x d x = \frac{1}{2} d u$ . Then

\begin{aligned} \int\frac{x^{3}}{\sqrt{x^{2}+9}}\,dx &= \int\frac{x^{2}}{\sqrt{x^{2}+9}}\,x\,dx\\ &= \frac{1}{2}\int\frac{u-9}{\sqrt{u}}\,du\\ &= \frac{1}{2}\int\left(u^{1/2}-9u^{-1/2}\right)\,du\\ &= \frac{1}{3}u^{3/2}-9u^{1/2}+C\\ &= \frac{1}{3}(x^{2}+9)^{3/2}-9\sqrt{x^{2}+9}+C. \end{aligned}

Trigonometric substitution. Let $x = 3 \tan θ$ , so $d x = 3 \sec^{2} θ d θ$ and $\sqrt{x^{2} + 9} = 3 \sec θ$ . Then

\begin{aligned} \int\frac{x^{3}}{\sqrt{x^{2}+9}}\,dx &= \int\frac{27\tan^{3}\theta\cdot3\sec^{2}\theta}{3\sec\theta}\,d\theta\\ &= 27\int\tan^{3}\theta\sec\theta\,d\theta. \end{aligned}

By Case 2 of Section 6.1 (the exponent of $\tan θ$ is odd), with $u = \sec θ$ ,

27 \int (\sec^{2} θ - 1) \sec θ \tan θ d θ = 27 (\frac{\sec^{3} θ}{3} - \sec θ) + C .

Since $\sec θ = \frac{\sqrt{x^{2} + 9}}{3}$ ,

27 (\frac{(x^{2} + 9)^{3 / 2}}{81} - \frac{\sqrt{x^{2} + 9}}{3}) + C = \frac{1}{3} (x^{2} + 9)^{3 / 2} - 9 \sqrt{x^{2} + 9} + C,

which agrees with the first method. The ordinary substitution is clearly faster here, because the odd power of $x$ supplies the differential directly.

Evaluate $\int \frac{d x}{x^{2} \sqrt{x^{2} - 9}}$ , $x > 3$ .

Solution

Let $x = 3 \sec θ$ with $0 < θ < \frac{π}{2}$ , so $d x = 3 \sec θ \tan θ d θ$ and $\sqrt{x^{2} - 9} = 3 \tan θ$ (positive on this interval). Then

\begin{aligned} \int\frac{dx}{x^{2}\sqrt{x^{2}-9}} &= \int\frac{3\sec\theta\tan\theta\,d\theta}{9\sec^{2}\theta\cdot3\tan\theta}\\ &= \frac{1}{9}\int\frac{d\theta}{\sec\theta}\\ &= \frac{1}{9}\int\cos\theta\,d\theta\\ &= \frac{\sin\theta}{9}+C. \end{aligned}

From $\sec θ = \frac{x}{3}$ we get $\cos θ = \frac{3}{x}$ and $\sin θ = \frac{\sqrt{x^{2} - 9}}{x}$ , so

\int \frac{d x}{x^{2} \sqrt{x^{2} - 9}} = \frac{\sqrt{x^{2} - 9}}{9 x} + C .

Evaluate $\int \frac{\sqrt{x^{2} - 4}}{x} d x$ , $x > 2$ .

Solution

Let $x = 2 \sec θ$ with $0 \leq θ < \frac{π}{2}$ , so $d x = 2 \sec θ \tan θ d θ$ and $\sqrt{x^{2} - 4} = 2 \tan θ$ . Then

\begin{aligned} \int\frac{\sqrt{x^{2}-4}}{x}\,dx &= \int\frac{2\tan\theta}{2\sec\theta}\cdot 2\sec\theta\tan\theta\,d\theta\\ &= 2\int\tan^{2}\theta\,d\theta\\ &= 2\int(\sec^{2}\theta-1)\,d\theta\\ &= 2\tan\theta-2\theta+C. \end{aligned}

Since $\tan θ = \frac{\sqrt{x^{2} - 4}}{2}$ and $\cos θ = \frac{2}{x}$ , so that $θ = \arccos \frac{2}{x}$ ,

\int \frac{\sqrt{x^{2} - 4}}{x} d x = \sqrt{x^{2} - 4} - 2 \arccos \frac{2}{x} + C .

Evaluate $\int \frac{d x}{(x^{2} - 4)^{3 / 2}}$ , $x > 2$ .

Solution

Let $x = 2 \sec θ$ with $0 < θ < \frac{π}{2}$ , so $d x = 2 \sec θ \tan θ d θ$ and $(x^{2} - 4)^{3 / 2} = 8 \tan^{3} θ$ . Then

\begin{aligned} \int\frac{dx}{(x^{2}-4)^{3/2}} &= \int\frac{2\sec\theta\tan\theta\,d\theta}{8\tan^{3}\theta}\\ &= \frac{1}{4}\int\frac{\sec\theta}{\tan^{2}\theta}\,d\theta\\ &= \frac{1}{4}\int\frac{\cos\theta}{\sin^{2}\theta}\,d\theta. \end{aligned}

With $w = \sin θ$ this is $\frac{1}{4} \int w^{- 2} d w = - \frac{1}{4 \sin θ} + C$ . Since $\sin θ = \frac{\sqrt{x^{2} - 4}}{x}$ ,

\int \frac{d x}{(x^{2} - 4)^{3 / 2}} = - \frac{x}{4 \sqrt{x^{2} - 4}} + C .

Evaluate $\int \frac{d x}{\sqrt{x^{2} - 2 x}}$ , $x > 2$ .

Solution

Complete the square: $x^{2} - 2 x = (x - 1)^{2} - 1$ . Put $u = x - 1$ , so $d u = d x$ and

\int \frac{d x}{\sqrt{x^{2} - 2 x}} = \int \frac{d u}{\sqrt{u^{2} - 1}} .

Let $u = \cosh t$ with $t > 0$ , so $d u = \sinh t d t$ and $\sqrt{u^{2} - 1} = \sinh t$ . Then

\int \frac{d u}{\sqrt{u^{2} - 1}} = \int d t = t + C = arccosh u + C = \ln (u + \sqrt{u^{2} - 1}) + C .

Therefore

\int \frac{d x}{\sqrt{x^{2} - 2 x}} = \ln (x - 1 + \sqrt{x^{2} - 2 x}) + C .

Evaluate $\int \frac{d x}{\sqrt{3 - 2 x - x^{2}}}$ .

Solution

Complete the square. Since $3 - 2 x - x^{2} = - (x^{2} + 2 x - 3) = - [(x + 1)^{2} - 4] = 4 - (x + 1)^{2}$ , put $u = x + 1$ :

\int \frac{d x}{\sqrt{3 - 2 x - x^{2}}} = \int \frac{d u}{\sqrt{4 - u^{2}}} .

Let $u = 2 \sin θ$ , so $d u = 2 \cos θ d θ$ and $\sqrt{4 - u^{2}} = 2 \cos θ$ :

\int \frac{d u}{\sqrt{4 - u^{2}}} = \int d θ = θ + C = \arcsin \frac{u}{2} + C .

Therefore

\int \frac{d x}{\sqrt{3 - 2 x - x^{2}}} = \arcsin \frac{x + 1}{2} + C .

Evaluate $\int e^{x} \sqrt{1 - e^{2 x}} d x$ .

Solution

First substitute $u = e^{x}$ , so $d u = e^{x} d x$ and $e^{2 x} = u^{2}$ :

\int e^{x} \sqrt{1 - e^{2 x}} d x = \int \sqrt{1 - u^{2}} d u .

By the first example of this section,

\int \sqrt{1 - u^{2}} d u = \frac{1}{2} \arcsin u + \frac{1}{2} u \sqrt{1 - u^{2}} + C .

Therefore

\int e^{x} \sqrt{1 - e^{2 x}} d x = \frac{1}{2} \arcsin (e^{x}) + \frac{1}{2} e^{x} \sqrt{1 - e^{2 x}} + C .

Evaluate $\int \frac{d x}{x^{2} + 2 x + 5}$ .

Solution

There is no radical here, but the same completing-the-square idea applies. Since $x^{2} + 2 x + 5 = (x + 1)^{2} + 4$ , put $u = x + 1$ :

\int \frac{d x}{x^{2} + 2 x + 5} = \int \frac{d u}{u^{2} + 4} .

Let $u = 2 \tan θ$ , so $d u = 2 \sec^{2} θ d θ$ and $u^{2} + 4 = 4 \sec^{2} θ$ :

\int \frac{d u}{u^{2} + 4} = \int \frac{2 \sec^{2} θ d θ}{4 \sec^{2} θ} = \frac{θ}{2} + C = \frac{1}{2} \arctan \frac{u}{2} + C .

Therefore

\int \frac{d x}{x^{2} + 2 x + 5} = \frac{1}{2} \arctan \frac{x + 1}{2} + C .

Integrals Involving the Radical √(a²−x²)

Integrals Involving the Radical √(a²+x²)

Integrals Involving the Radical √(x²−a²)

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