Integrals Of Trigonometric Functions

Every trigonometric integral in this section is solved with only two tools: a trigonometric identity that reshapes the integrand, followed by a substitution that finishes the job. The art lies entirely in choosing the right identity.

Integral	Result
$\int \sin x d x$	$- \cos x + C$
$\int \cos x d x$	$\sin x + C$
$\int \sec^{2} x d x$	$\tan x + C$
$\int \csc^{2} x d x$	$- \cot x + C$
$\int \sec x \tan x d x$	$\sec x + C$
$\int \csc x \cot x d x$	$- \csc x + C$
$\int \tan x d x$	$\ln < / t d >< / t r >< t r >< t d >$ \displaystyle\int\cot x\,dx $< / t d >< t d >$ \ln
$\int \sec x d x$	$\ln < / t d >< / t r >< t r >< t d >$ \displaystyle\int\csc x\,dx $< / t d >< t d >$ -\ln

Situation	Strategy
$\sin^{m} x \cos^{n} x$ , $m$ odd	Peel off one $\sin x$ for $d u$ ; convert the rest with $\sin^{2} x = 1 - \cos^{2} x$ ; let $u = \cos x$
$\sin^{m} x \cos^{n} x$ , $n$ odd	Peel off one $\cos x$ for $d u$ ; convert the rest with $\cos^{2} x = 1 - \sin^{2} x$ ; let $u = \sin x$
$\sin^{m} x \cos^{n} x$ , both even	Use the half-angle identities repeatedly
$\tan^{m} x \sec^{n} x$ , $n$ even	Peel off $\sec^{2} x$ for $d u$ ; convert the rest with $\sec^{2} x = 1 + \tan^{2} x$ ; let $u = \tan x$
$\tan^{m} x \sec^{n} x$ , $m$ odd	Peel off $\sec x \tan x$ for $d u$ ; convert the rest with $\tan^{2} x = \sec^{2} x - 1$ ; let $u = \sec x$
$\sin a x \cos b x$ , $\sin a x \sin b x$ , $\cos a x \cos b x$	Use the product-to-sum identities

Integrals That Follow Directly From Derivatives

In the previous chapter we saw

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\begin{aligned} \int\sin x\ dx &= -\cos x+C\\ \int\cos x\ dx &= \sin x+C. \end{aligned}}

In this section, we will investigate the integrals of the other trigonometric functions.

What is $\int \sec^{2} x d x$ ? Because (\tan x)'=\sec^{2}x, we have

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int\sec^{2}x\ dx=\tan x+C.}

Similarly, because (\cot x)'=-\csc^{2}x,

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int\csc^{2}x\ dx=-\cot x+C.}

In words: every derivative formula for a trigonometric function, read backwards, is an integration formula. The remaining trigonometric functions, $\tan x$ , $\cot x$ , $\sec x$ , and $\csc x$ , are not the derivative of any function we have named, so their integrals take a little more work.

The Integrals of tan x and cot x

It is worth repeating an important integral that we evaluated in the section on integration by substitution.

Find $\int \tan x d x$ .

Solution

Because $\tan x = \sin x / \cos x$ and (\cos x)'=-\sin x, let $u = \cos x$ . Then

d u = - \sin x d x,

and

\begin{aligned} \int\tan x\ dx &= \int\frac{\sin x}{\cos x}\,dx\\ &= \int\frac{-du}{u}\\ &= -\ln|u|+C\\ &= -\ln|\cos x|+C. \end{aligned}

Recall that $\ln A^{B} = B \ln A$ . So

\begin{aligned} -\ln|\cos x|+C &= \ln\left(|\cos x|^{-1}\right)+C\\ &= \ln|\sec x|+C. &&{\small(\sec x=\tfrac{1}{\cos x})} \end{aligned}

Therefore, the result can also be written as

\int \tan x d x = \ln | \sec x | + C .

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int\tan x\ dx=\ln|\sec x|+C=-\ln|\cos x|+C.}

Find $\int \cot x d x$ .

Solution

Because $\cot x = \cos x / \sin x$ , if we let $u = \sin x$ , then

d u = \cos x d x

and

\begin{aligned} \int\cot x\ dx &= \int\frac{\cos x}{\sin x}\,dx\\ &= \int\frac{du}{u}\\ &= \ln|u|+C\\ &= \ln|\sin x|+C. \end{aligned}

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int\cot x\ dx=\ln|\sin x|+C.}

Notice the pattern that makes both of these work: in each case the numerator is, up to sign, the derivative of the denominator. Any integral of the form \displaystyle\int\frac{g'(x)}{g(x)}\,dx equals $\ln | g (x) | + C$ .

The Integral of sec x tan x

Find $\int \sec x \tan x d x$ .

Solution

Because

\frac{d}{d x} \sec x = \sec x \tan x,

we already know that the result of this integral is $\sec x + C$ . However, even if we did not know the result, we could use integration by substitution to derive it:

\begin{aligned} \int\sec x\tan x\ dx &= \int\frac{1}{\cos x}\frac{\sin x}{\cos x}\,dx\\ &= \int\frac{\sin x}{\cos^{2}x}\,dx. \end{aligned}

Let $u = \cos x$ , then

d u = - \sin x d x .

\begin{aligned} \int\frac{\sin x}{\cos^{2}x}\,dx &= \int-\frac{du}{u^{2}}\\ &= -\int u^{-2}\,du\\ &= -(-u^{-1})+C\\ &= \frac{1}{u}+C\\ &= \frac{1}{\cos x}+C &&{\small(u=\cos x)}\\ &= \sec x+C. \end{aligned}

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int \sec x \tan x\;dx=\sec x+C}

Find $\int e^{x} \sec (e^{x}) \tan (e^{x}) d x$ .

Solution

Let $u = e^{x}$ . Then

d u = e^{x} d x

and

\begin{aligned} \int\sec(\underbrace{e^{x}}_{u})\tan(\underbrace{e^{x}}_{u})\underbrace{e^{x}\,dx}_{du} &= \int\sec u\tan u\ du\\ &= \sec u+C\\ &= \sec e^{x}+C. \end{aligned}

The Integrals of sec x and csc x

Find $\int \sec x d x$ .

Solution

There are several ways to evaluate this integral; each involves a trick. The easiest way is to multiply and divide the integrand by $\sec x + \tan x$ :

\begin{aligned} \int\sec x\ dx &= \int\sec x\frac{\sec x+\tan x}{\sec x+\tan x}\,dx\\ &= \int\frac{\sec^{2}x+\sec x\tan x}{\sec x+\tan x}\,dx. \end{aligned}

Let $u = \sec x + \tan x$ , which gives

d u = (\sec x \tan x + \sec^{2} x) d x .

The numerator is now exactly $d u$ . Substituting these into the integral yields

\begin{aligned} \int\frac{\sec^{2}x+\sec x\tan x}{\sec x+\tan x}\,dx &= \int\frac{du}{u}\\ &= \ln|u|+C\\ &= \ln|\sec x+\tan x|+C. \end{aligned}

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int\sec x\ dx=\ln|\sec x+\tan x|+C.}

Find $\int \csc x d x$ .

Solution

Similar to the evaluation of $\int \sec x d x$ , we can integrate $\csc x$ by multiplying and dividing by $(\csc x + \cot x)$ :

\int \csc x d x = \int \csc x \cdot \frac{\csc x + \cot x}{\csc x + \cot x} d x

\int \csc x d x = \int \frac{\csc^{2} x + \csc x \cot x}{\csc x + \cot x} d x

Now let

u = \csc x + \cot x .

Then the differential is

d u = (- \csc x \cot x - \csc^{2} x) d x .

Factoring out the negative sign gives

d u = - (\csc^{2} x + \csc x \cot x) d x

- d u = (\csc^{2} x + \csc x \cot x) d x .

Substituting $u$ and $d u$ back into the integral, we obtain

\begin{aligned} \int \csc x\; dx &= \int \frac{-du}{u} \\ &= -\int \frac{1}{u} \, du\\ &= -\ln|u| + C. \end{aligned}

Finally substituting back for $u$ yields

\int \csc x d x = - \ln | \csc x + \cot x | + C .

Alternative Form. Using the logarithmic property $- \ln A = \ln \frac{1}{A}$ :

- \ln | \csc x + \cot x | = \ln | \frac{1}{\csc x + \cot x} | .

If we multiply the numerator and denominator inside the absolute value by $(\csc x - \cot x)$ and apply the trigonometric identity $\csc^{2} x - \cot^{2} x = 1$ , we get

\ln | \frac{\csc x - \cot x}{\csc^{2} x - \cot^{2} x} | = \ln | \csc x - \cot x | .

Both $- \ln | \csc x + \cot x | + C$ and $\ln | \csc x - \cot x | + C$ are correct forms.

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\int \csc x\; dx=-\ln|\csc x+\cot x|+C}

The Integrals of sin²x and cos²x

Neither $\sin^{2} x$ nor $\cos^{2} x$ is the derivative of anything obvious, and no substitution helps directly. The way out is to lower the power first, using the half-angle identities

\sin^{2} x = \frac{1 - \cos 2 x}{2}, \cos^{2} x = \frac{1 + \cos 2 x}{2} .

Each identity trades a squared function of $x$ for a first power of a function of $2 x$ , and first powers we can integrate.

Find $\int \sin^{2} x d x$ .

Solution

We lower the power with the half-angle identity for the sine, $\sin^{2} θ = \frac{1 - \cos 2 θ}{2}$ , where $θ$ is the angle inside the sine. Taking $θ = x$ gives $\sin^{2} x = \frac{1 - \cos 2 x}{2}$ , so

\begin{aligned} \int\sin^{2}x\ dx &= \frac{1}{2}\int(1-\cos2x)\,dx\\ &= \frac{1}{2}x-\frac{1}{4}\sin2x+C, \end{aligned}

where $\int \cos 2 x d x$ is evaluated using the substitution $u = 2 x$ .

Find $\int \cos^{2} x d x$ .

Solution

We lower the power with the half-angle identity for the cosine, $\cos^{2} θ = \frac{1 + \cos 2 θ}{2}$ , where $θ$ is the angle inside the cosine. Taking $θ = x$ gives $\cos^{2} x = \frac{1 + \cos 2 x}{2}$ , so

\begin{aligned} \int\cos^{2}x\ dx &= \frac{1}{2}\int(1+\cos2x)\,dx\\ &= \frac{1}{2}x+\frac{1}{4}\sin2x+C. \end{aligned}

\int \sin^{2} x d x = \frac{x}{2} - \frac{\sin 2 x}{4} + C, \int \cos^{2} x d x = \frac{x}{2} + \frac{\sin 2 x}{4} + C .

Notice that adding these two results gives $x + C$ , exactly as it must, because $\sin^{2} x + \cos^{2} x = 1$ . This is a useful way to check that you have the signs right.

Products of Sines and Cosines of Different Angles

In many applied problems such as mechanical vibrations and electromagnetic waves, we encounter the following trigonometric integrals

\int \sin a x \cos b x d x, \int \sin a x \sin b x d x, \int \cos a x \cos b x d x .

To evaluate these integrals, we use the product-to-sum identities

\sin a x \cos b x = \frac{1}{2} [\sin (a - b) x + \sin (a + b) x]

\sin a x \sin b x = \frac{1}{2} [\cos (a - b) x - \cos (a + b) x]

\cos a x \cos b x = \frac{1}{2} [\cos (a - b) x + \cos (a + b) x]

In each of these identities, a is the coefficient of $x$ inside the first factor and b is the coefficient of $x$ inside the second factor. For instance, in $\sin 3 x \cos 5 x$ the first factor is $\sin 3 x$ , so $a = 3$ , and the second factor is $\cos 5 x$ , so $b = 5$ . The two angles appearing on the right are then $(a - b) x = - 2 x$ and $(a + b) x = 8 x$ .

Each identity replaces a product, which we cannot integrate directly, by a sum of two single trigonometric functions, each of which integrates in one step by the linear substitution rule.

Find $\int \sin 3 x \cos 5 x d x$ .

Solution

The integrand is a sine times a cosine of a different angle, so we use the product-to-sum identity

\sin a x \cos b x = \frac{1}{2} [\sin (a - b) x + \sin (a + b) x],

in which a is the coefficient of $x$ inside the sine and b is the coefficient of $x$ inside the cosine. Our integrand is $\sin 3 x \cos 5 x$ , so the sine carries $3 x$ and the cosine carries $5 x$ , giving

a = 3, b = 5, (a - b) x = - 2 x, (a + b) x = 8 x .

Since the sine is an odd function, $\sin (- 2 x) = - \sin 2 x$ , so

\begin{aligned} \int\sin3x\,\cos5x\,dx &= \frac{1}{2}\int\left[\sin(-2x)+\sin8x\right]\,dx\\ &= \frac{1}{2}\int\left[\sin8x-\sin2x\right]\ dx\\ &= \frac{1}{2}\left(-\frac{1}{8}\cos8x+\frac{1}{2}\cos2x\right)+C\\ &= \frac{1}{4}\cos2x-\frac{1}{16}\cos8x+C. \end{aligned}

Two cautions when applying the product-to-sum identities, in which a is the coefficient of $x$ in the first factor and b is the coefficient of $x$ in the second. First, $a - b$ may be negative; use $\sin (- θ) = - \sin θ$ and $\cos (- θ) = \cos θ$ to clean up the result, as in the example above. Second, if $a = b$ then $(a - b) x = 0$ , so the term $\cos (a - b) x$ becomes $\cos 0 = 1$ , and the identity reduces to the half-angle formulas of the previous subsection.

Integrating sin^m x cosⁿ x When Either m or n Is a Positive Odd Integer

When $\sin x$ has an odd exponent, we first reduce the integral to the form

\int (terms involving only \cos x) \sin x d x,

and when $\cos x$ has an odd exponent, we reduce the integral to the form

\int (terms involving only \sin x) \cos x d x .

Then we can perform the integration by means of

\int u^{n} d u = \frac{u^{n + 1}}{n + 1} + C .

The reason this works is worth stating plainly: an odd power always leaves one factor over after we pair the rest into squares, and that leftover factor is exactly the differential we need. The following examples will illustrate how to use this.

Find $\int \sin^{2} x \cos^{5} x d x$ .

Solution

Here the exponent of $\cos x$ is odd, so we separate one factor of $\cos x$ for the differential and convert the remaining $\cos^{4} x$ into sines with the Pythagorean identity

\sin^{2} x + \cos^{2} x = 1, that is, \cos^{2} x = 1 - \sin^{2} x,

so that $\cos^{4} x = {(\cos^{2} x)}^{2} = (1 - \sin^{2} x)^{2}$ . Thus

\begin{aligned} \int\sin^{2}x\cos^{5}x\, dx &= \int\sin^{2}x\cos^{4}x\cos x\,dx\\ &= \int\sin^{2}x\,(1-\sin^{2}x)^{2}\cos x\,dx\\ &= \int(\sin^{2}x-2\sin^{4}x+\sin^{6}x)\cos x\,dx\\ &= \int\sin^{2}x\cos x\ dx-2\int\sin^{4}x\cos x\ dx\\ &\qquad +\int\sin^{6}x\cos x\ dx. \end{aligned}

Now let

u = \sin x \Rightarrow d u = \cos x d x .

Thus

\begin{aligned} \int\sin^{2}x\cos^{5}x\,dx=\int&\underbrace{\sin^{2}x}_{u^{2}}\underbrace{\cos x\ dx}_{du}-2\int\underbrace{\sin^{4}x}_{u^{4}}\underbrace{\cos x\ dx}_{du}\\ &+\int\underbrace{\sin^{6}x}_{u^{6}}\underbrace{\cos x\ dx}_{du} \end{aligned}

\int \sin^{2} x \cos^{5} x d x = \frac{\sin^{3} x}{3} - 2 \frac{\sin^{5} x}{5} + \frac{\sin^{7} x}{7} + C .

Find $\int \cos^{3} x d x$ .

Solution

The exponent of $\cos x$ is odd, so we split off one factor of $\cos x$ for the differential and convert the remaining $\cos^{2} x$ into sines by the Pythagorean identity

\sin^{2} x + \cos^{2} x = 1, that is, \cos^{2} x = 1 - \sin^{2} x .

Therefore

\begin{aligned} \int\cos^{3}x\,dx &= \int\cos^{2}x\cos x\,dx\\ &= \int(1-\sin^{2}x)\cos x\,dx\\ &= \int\cos x\,dx-\int\underbrace{\sin^{2}x}_{u^{2}}\ \underbrace{\cos x\,dx}_{du}\\ &= \sin x-\underbrace{\frac{\sin^{3}x}{3}}_{u^{3}/3}+C. \end{aligned}

Find $\int \frac{\cos^{5} x}{\sqrt{\sin x}} d x$ .

Solution

Writing the integrand as $\sin^{m} x \cos^{n} x$ , we have $m = - 1 / 2$ and $n = 5$ , since $\frac{1}{\sqrt{\sin x}} = \sin^{- 1 / 2} x$ . Notice that the strategy does not require $m$ to be a positive integer; only the odd exponent on $\cos x$ matters. We separate a factor of $\cos x$ from $\cos^{5} x$ and express the remaining factor $\cos^{4} x$ in terms of $\sin x$ using the Pythagorean identity

\sin^{2} x + \cos^{2} x = 1, that is, \cos^{2} x = 1 - \sin^{2} x,

so that $\cos^{4} x = (1 - \sin^{2} x)^{2}$ . That is,

\begin{aligned} \int\frac{\cos^{5}x}{\sqrt{\sin x}}\,dx &= \int\frac{\cos^{4}x}{\sqrt{\sin x}}\cos x\ dx\\ &= \int\frac{(1-\sin^{2}x)^{2}}{\sqrt{\sin x}}\cos x\ dx. \end{aligned}

Now let $u = \sin x$ . Then $d u = \cos x d x$ and

\begin{aligned} \int\frac{(1-\sin^{2}x)^{2}}{\sqrt{\sin x}}\cos x\ dx &= \int\frac{(1-u^{2})^{2}}{\sqrt{u}}\,du\\ &= \int\frac{1-2u^{2}+u^{4}}{\sqrt{u}}\,du\\ &= \int\left(u^{-1/2}-2u^{3/2}+u^{7/2}\right)\,du\\ &= 2u^{1/2}-\frac{4}{5}u^{5/2}+\frac{2}{9}u^{9/2}+C\\ &= 2\sqrt{\sin x}-\frac{4}{5}\sqrt{\sin^{5}x}+\frac{2}{9}\sqrt{\sin^{9}x}+C. \end{aligned}

Integrating sin^m x cosⁿ x When Both m and n Are Positive Even Integers

When both exponents are even, the odd-power strategy no longer applies: there is no spare factor to serve as $d u$ . Instead, we use the half-angle identities

\sin^{2} x = \frac{1 - \cos 2 x}{2}, \cos^{2} x = \frac{1 + \cos 2 x}{2},

and the double-angle identity

\sin x \cos x = \frac{1}{2} \sin 2 x

repeatedly until no even powers remain. The resulting integrals involve only first powers of cosines of multiple angles, which are straightforward to integrate.

Find $\int \sin^{2} x \cos^{2} x d x$ .

Solution

We use the double-angle identity for the sine,

\sin 2 θ = 2 \sin θ \cos θ, that is, \sin θ \cos θ = \frac{1}{2} \sin 2 θ,

in which $θ$ is the common angle of the sine and cosine. Here $θ = x$ , so $\sin x \cos x = \frac{1}{2} \sin 2 x$ and

\begin{aligned} \int\sin^{2}x\cos^{2}x\,dx &= \int\left(\sin x\cos x\right)^{2}\,dx\\ &= \int\left(\frac{1}{2}\sin2x\right)^{2}\,dx\\ &= \frac{1}{4}\int\sin^{2}2x\,dx. \end{aligned}

Now we apply $\sin^{2} 2 x = \frac{1 - \cos 4 x}{2}$ :

\begin{aligned} \frac{1}{4}\int\sin^{2}2x\,dx &= \frac{1}{4}\int\frac{1-\cos4x}{2}\,dx\\ &= \frac{1}{8}\int\left(1-\cos4x\right)\,dx\\ &= \frac{1}{8}\left(x-\frac{1}{4}\sin4x\right)+C\\ &= \frac{x}{8}-\frac{\sin4x}{32}+C. \end{aligned}

Find $\int \sin^{4} x \cos^{2} x d x$ .

Solution

We write $\sin^{4} x = (\sin^{2} x)^{2}$ and apply the half-angle identities

\sin^{2} θ = \frac{1 - \cos 2 θ}{2}, \cos^{2} θ = \frac{1 + \cos 2 θ}{2},

in which $θ$ is the angle inside the sine or cosine and $2 θ$ is twice that angle. Both factors here have $θ = x$ , so $2 θ = 2 x$ and

\sin^{2} x = \frac{1 - \cos 2 x}{2}, \cos^{2} x = \frac{1 + \cos 2 x}{2} .

Thus

\begin{aligned} \int\sin^{4}x\cos^{2}x\,dx &= \int\left(\frac{1-\cos2x}{2}\right)^{2}\cdot\frac{1+\cos2x}{2}\,dx\\ &= \frac{1}{8}\int(1-\cos2x)^{2}(1+\cos2x)\,dx. \end{aligned}

Expanding $(1 - \cos 2 x)^{2} (1 + \cos 2 x)$ :

\begin{aligned} (1-\cos2x)^{2}(1+\cos2x) &= (1-2\cos2x+\cos^{2}2x)(1+\cos2x)\\ &= 1+\cos2x-2\cos2x-2\cos^{2}2x\\ &\qquad+\cos^{2}2x+\cos^{3}2x\\ &= 1-\cos2x-\cos^{2}2x+\cos^{3}2x. \end{aligned}

For the $\cos^{2} 2 x$ term we again use the half-angle identity $\cos^{2} θ = \frac{1 + \cos 2 θ}{2}$ , now with $θ = 2 x$ so that $2 θ = 4 x$ , giving $\cos^{2} 2 x = \frac{1 + \cos 4 x}{2}$ . For the $\cos^{3} 2 x$ term the exponent is odd, so we split off one factor of $\cos 2 x$ and convert the rest with the Pythagorean identity $\cos^{2} 2 x = 1 - \sin^{2} 2 x$ :

\cos^{3} 2 x = (1 - \sin^{2} 2 x) \cos 2 x .

Substituting back:

\begin{aligned} \int\sin^{4}x\cos^{2}x\,dx &= \frac{1}{8}\int\left[1-\cos2x-\frac{1+\cos4x}{2} +(1-\sin^{2}2x)\cos2x\right]\,dx\\ &= \frac{1}{8}\int\left[\frac{1}{2}-\frac{\cos4x}{2} -\sin^{2}2x\cos2x\right]\,dx. \end{aligned}

Integrating term by term, using $u = 2 x$ for $\cos 4 x$ and $u = \sin 2 x$ for the last term:

\begin{aligned} \int\sin^{4}x\cos^{2}x\,dx &= \frac{1}{8}\left[\frac{x}{2}-\frac{\sin4x}{8} -\frac{\sin^{3}2x}{6}\right]+C\\ &= \frac{x}{16}-\frac{\sin4x}{64}-\frac{\sin^{3}2x}{48}+C. \end{aligned}

Find $\int \cos^{4} x d x$ .

Solution

We write $\cos^{4} x = (\cos^{2} x)^{2}$ and apply the half-angle identity for the cosine,

\cos^{2} θ = \frac{1 + \cos 2 θ}{2},

where $θ$ is the angle inside the cosine. With $θ = x$ this reads $\cos^{2} x = \frac{1 + \cos 2 x}{2}$ , so

\begin{aligned} \int\cos^{4}x\,dx &= \int\left(\frac{1+\cos2x}{2}\right)^{2}\,dx\\ &= \frac{1}{4}\int\left(1+2\cos2x+\cos^{2}2x\right)\,dx. \end{aligned}

For the remaining $\cos^{2} 2 x$ term, we apply the same identity $\cos^{2} θ = \frac{1 + \cos 2 θ}{2}$ once more, this time with $θ = 2 x$ , so that $2 θ = 4 x$ :

\cos^{2} 2 x = \frac{1 + \cos 4 x}{2} .

Substituting back:

\begin{aligned} \frac{1}{4}\int&\left(1+2\cos2x+\cos^{2}2x\right)\,dx\\ &= \frac{1}{4}\int\left(1+2\cos2x+\frac{1+\cos4x}{2}\right)\,dx\\ &= \frac{1}{4}\int\left(\frac{3}{2}+2\cos2x+\frac{\cos4x}{2}\right)\,dx\\ &= \frac{1}{4}\left(\frac{3}{2}x+\sin2x+\frac{\sin4x}{8}\right)+C\\ &= \frac{3x}{8}+\frac{\sin2x}{4}+\frac{\sin4x}{32}+C. \end{aligned}

Integrating tan^m x secⁿ x When m Is a Positive Odd Integer or n Is a Positive Even Integer

The key identities for this family of integrals are

\sec^{2} x = 1 + \tan^{2} x and \frac{d}{d x} \tan x = \sec^{2} x, \frac{d}{d x} \sec x = \sec x \tan x .

Two cases arise naturally.

Case 1: $n$ is a positive even integer. The goal is to use the substitution

u = \tan x, d u = \sec^{2} x d x .

For this to work, we need a bare factor of $\sec^{2} x$ to serve as the differential $d u$ . Since $n$ is even, we can always peel one $\sec^{2} x$ off the integrand and reserve it for $d u$ . The remaining factor $\sec^{n - 2} x$ has an even exponent, so it can be converted entirely into powers of $\tan x$ using

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

repeatedly. The integral then becomes a polynomial in $u = \tan x$ :

\begin{aligned} \int\tan^{m}x\,\sec^{n}x\,dx &= \int\tan^{m}x\,\sec^{n-2}x\cdot\sec^{2}x\,dx\\ &= \int u^{m}(\text{polynomial in }u)\,du. \end{aligned}

Case 2: $m$ is a positive odd integer. The goal is to use the substitution

u = \sec x, d u = \sec x \tan x d x .

For this to work, we need a bare factor of $\sec x \tan x$ to serve as the differential $d u$ . Since $m$ is odd, we can always peel one factor of $\tan x$ off $\tan^{m} x$ and pair it with one factor of $\sec x$ from $\sec^{n} x$ , reserving the product $\sec x \tan x$ for $d u$ . The remaining factor $\tan^{m - 1} x$ has an even exponent, so it can be converted entirely into powers of $\sec x$ using $\tan^{2} x = \sec^{2} x - 1$ repeatedly. The integral then becomes a polynomial in $u = \sec x$ :

\begin{aligned} \int\tan^{m}x\,\sec^{n}x\,dx &= \int\tan^{m-1}x\,\sec^{n-1}x\cdot\sec x\tan x\,dx\\ &= \int u^{n-1}(\text{polynomial in }u)\,du. \end{aligned}

Find $\int \tan^{2} x \sec^{4} x d x$ .

Solution

The exponent of $\sec x$ is even, so we apply Case 1. We separate one factor of $\sec^{2} x$ to serve as the differential, and convert the remaining $\sec^{2} x$ into tangents by the Pythagorean identity

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

Therefore

\begin{aligned} \int\tan^{2}x\sec^{4}x\,dx &= \int\tan^{2}x\sec^{2}x\cdot\sec^{2}x\,dx\\ &= \int\tan^{2}x\,(1+\tan^{2}x)\sec^{2}x\,dx. \end{aligned}

Let $u = \tan x$ , so $d u = \sec^{2} x d x$ :

\begin{aligned} \int u^{2}(1+u^{2})\,du &= \int\left(u^{2}+u^{4}\right)\,du\\ &= \frac{u^{3}}{3}+\frac{u^{5}}{5}+C\\ &= \frac{\tan^{3}x}{3}+\frac{\tan^{5}x}{5}+C. \end{aligned}

Find $\int \tan^{3} x \sec^{3} x d x$ .

Solution

The exponent of $\tan x$ is odd, so we apply Case 2. We separate a factor of $\sec x \tan x$ to serve as the differential, and convert the remaining $\tan^{2} x$ into secants by the Pythagorean identity

\sec^{2} x = 1 + \tan^{2} x, that is, \tan^{2} x = \sec^{2} x - 1.

Therefore

\begin{aligned} \int\tan^{3}x\sec^{3}x\,dx &= \int\tan^{2}x\sec^{2}x\cdot\sec x\tan x\,dx\\ &= \int(\sec^{2}x-1)\sec^{2}x\cdot\sec x\tan x\,dx. \end{aligned}

Let $u = \sec x$ , so $d u = \sec x \tan x d x$ :

\begin{aligned} \int(u^{2}-1)u^{2}\,du &= \int\left(u^{4}-u^{2}\right)\,du\\ &= \frac{u^{5}}{5}-\frac{u^{3}}{3}+C\\ &= \frac{\sec^{5}x}{5}-\frac{\sec^{3}x}{3}+C. \end{aligned}

Find $\int \sqrt{\tan x} \sec^{4} x d x$ .

Solution

We can write this integral as $\int \tan^{1 / 2} x \sec^{4} x d x$ . The exponent of $\sec x$ is even, so we apply Case 1. Notice again that the exponent on $\tan x$ need not be an integer. We separate one factor of $\sec^{2} x$ to serve as the differential, and convert the remaining $\sec^{2} x$ into tangents by the Pythagorean identity

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

Therefore

\begin{aligned} \int\tan^{1/2}x\,\sec^{4}x\,dx &= \int\tan^{1/2}x\,\sec^{2}x\cdot\sec^{2}x\,dx\\ &= \int\tan^{1/2}x\,(1+\tan^{2}x)\sec^{2}x\,dx. \end{aligned}

Let $u = \tan x$ , so $d u = \sec^{2} x d x$ :

\begin{aligned} \int u^{1/2}(1+u^{2})\,du &= \int\left(u^{1/2}+u^{5/2}\right)\,du\\ &= \frac{2}{3}u^{3/2}+\frac{2}{7}u^{7/2}+C\\ &= \frac{2}{3}\tan^{3/2}x+\frac{2}{7}\tan^{7/2}x+C. \end{aligned}

Exercises

Each of the following can be done with the basic formulas, the substitution rule, and the identities of this section. No other technique is needed.

Evaluate $\int \tan 3 x d x$ .

Solution

Let $u = 3 x$ , so $d u = 3 d x$ and $d x = \frac{1}{3} d u$ . Then

\begin{aligned} \int\tan 3x\,dx &= \frac{1}{3}\int\tan u\,du\\ &= \frac{1}{3}\ln|\sec u|+C\\ &= \frac{1}{3}\ln|\sec 3x|+C. \end{aligned}

Equivalently, $- \frac{1}{3} \ln | \cos 3 x | + C$ .

Evaluate $\int x \cot (x^{2}) d x$ .

Solution

Let $u = x^{2}$ . Then $d u = 2 x d x$ , so $x d x = \frac{1}{2} d u$ and

\begin{aligned} \int x\cot(x^{2})\,dx &= \frac{1}{2}\int\cot u\,du\\ &= \frac{1}{2}\ln|\sin u|+C\\ &= \frac{1}{2}\ln|\sin(x^{2})|+C. \end{aligned}

Evaluate $\int \csc 2 x d x$ .

Solution

Let $u = 2 x$ , so $d x = \frac{1}{2} d u$ . Using the formula for $\int \csc u d u$ established above,

\begin{aligned} \int\csc 2x\,dx &= \frac{1}{2}\int\csc u\,du\\ &= -\frac{1}{2}\ln|\csc u+\cot u|+C\\ &= -\frac{1}{2}\ln|\csc 2x+\cot 2x|+C. \end{aligned}

Evaluate $\int \sin^{3} x d x$ .

Solution

The exponent of $\sin x$ is odd, so we peel off one factor of $\sin x$ to serve as the differential and convert what remains into cosines. The conversion uses the Pythagorean identity

\sin^{2} x + \cos^{2} x = 1, that is, \sin^{2} x = 1 - \cos^{2} x .

Therefore

\begin{aligned} \int\sin^{3}x\,dx &= \int\sin^{2}x\,\sin x\,dx\\ &= \int(1-\cos^{2}x)\sin x\,dx. \end{aligned}

Let $u = \cos x$ , so $d u = - \sin x d x$ :

\begin{aligned} \int(1-u^{2})(-du) &= \int(u^{2}-1)\,du\\ &= \frac{u^{3}}{3}-u+C\\ &= \frac{\cos^{3}x}{3}-\cos x+C. \end{aligned}

Evaluate $\int \cos^{5} x d x$ .

Solution

The exponent of $\cos x$ is odd, so we peel off one factor of $\cos x$ for the differential. The remaining $\cos^{4} x$ is converted into sines by the Pythagorean identity

\sin^{2} x + \cos^{2} x = 1, that is, \cos^{2} x = 1 - \sin^{2} x,

which squared gives $\cos^{4} x = {(\cos^{2} x)}^{2} = (1 - \sin^{2} x)^{2}$ . Therefore

\begin{aligned} \int\cos^{5}x\,dx &= \int(1-\sin^{2}x)^{2}\cos x\,dx. \end{aligned}

Let $u = \sin x$ , so $d u = \cos x d x$ :

\begin{aligned} \int(1-u^{2})^{2}\,du &= \int\left(1-2u^{2}+u^{4}\right)\,du\\ &= u-\frac{2u^{3}}{3}+\frac{u^{5}}{5}+C\\ &= \sin x-\frac{2\sin^{3}x}{3}+\frac{\sin^{5}x}{5}+C. \end{aligned}

Evaluate $\int \sin^{3} x \cos^{3} x d x$ .

Solution

Both exponents are odd, so either factor may supply the differential. We peel off a factor of $\cos x$ and convert the remaining $\cos^{2} x$ using the Pythagorean identity

\sin^{2} x + \cos^{2} x = 1, that is, \cos^{2} x = 1 - \sin^{2} x .

Therefore

\begin{aligned} \int\sin^{3}x\cos^{3}x\,dx &= \int\sin^{3}x\,(1-\sin^{2}x)\cos x\,dx. \end{aligned}

Let $u = \sin x$ , so $d u = \cos x d x$ :

\begin{aligned} \int u^{3}(1-u^{2})\,du &= \int\left(u^{3}-u^{5}\right)\,du\\ &= \frac{u^{4}}{4}-\frac{u^{6}}{6}+C\\ &= \frac{\sin^{4}x}{4}-\frac{\sin^{6}x}{6}+C. \end{aligned}

Peeling off a factor of $\sin x$ instead gives $- \frac{\cos^{4} x}{4} + \frac{\cos^{6} x}{6} + C$ , which differs from the first answer only by a constant.

Evaluate $\int \sin^{2} 3 x d x$ .

Solution

Both exponents are even (here $\sin^{2} 3 x \cos^{0} 3 x$ ), so we lower the power with the half-angle identity for the sine,

\sin^{2} θ = \frac{1 - \cos 2 θ}{2} .

In this identity $θ$ stands for whatever angle appears inside the sine, and $2 θ$ is twice that angle. Our angle is $θ = 3 x$ , so $2 θ = 6 x$ and the identity reads

\sin^{2} 3 x = \frac{1 - \cos 6 x}{2} .

Therefore

\begin{aligned} \int\sin^{2}3x\,dx &= \frac{1}{2}\int(1-\cos 6x)\,dx\\ &= \frac{x}{2}-\frac{\sin 6x}{12}+C. \end{aligned}

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

Solution

We lower the power with the half-angle identity for the cosine,

\cos^{2} θ = \frac{1 + \cos 2 θ}{2},

in which $θ$ is the angle inside the cosine and $2 θ$ is twice that angle. Here $θ = \frac{x}{2}$ , so $2 θ = x$ and the identity reads

\cos^{2} \frac{x}{2} = \frac{1 + \cos x}{2} .

Therefore

\begin{aligned} \int\cos^{2}\frac{x}{2}\,dx &= \frac{1}{2}\int(1+\cos x)\,dx\\ &= \frac{x}{2}+\frac{\sin x}{2}+C. \end{aligned}

Evaluate $\int \sin^{4} x d x$ .

Solution

Write $\sin^{4} x = (\sin^{2} x)^{2}$ and apply the half-angle identity for the sine,

\sin^{2} θ = \frac{1 - \cos 2 θ}{2},

where $θ$ is the angle inside the sine. Here $θ = x$ , so $2 θ = 2 x$ and $\sin^{2} x = \frac{1 - \cos 2 x}{2}$ . Squaring,

\begin{aligned} \sin^{4}x &= \left(\frac{1-\cos 2x}{2}\right)^{2}\\ &= \frac{1-2\cos 2x+\cos^{2}2x}{4}. \end{aligned}

The term $\cos^{2} 2 x$ is still an even power, so we apply the half-angle identity for the cosine,

\cos^{2} θ = \frac{1 + \cos 2 θ}{2},

this time with $θ = 2 x$ , so that $2 θ = 4 x$ and $\cos^{2} 2 x = \frac{1 + \cos 4 x}{2}$ . Substituting,

\begin{aligned} \sin^{4}x &= \frac{1}{4}\left(1-2\cos 2x+\frac{1+\cos 4x}{2}\right)\\ &= \frac{1}{4}\left(\frac{3}{2}-2\cos 2x+\frac{\cos 4x}{2}\right). \end{aligned}

Integrating term by term,

\begin{aligned} \int\sin^{4}x\,dx &= \frac{1}{4}\left(\frac{3x}{2}-\sin 2x+\frac{\sin 4x}{8}\right)+C\\ &= \frac{3x}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C. \end{aligned}

Evaluate $\int \sin 4 x \cos 2 x d x$ .

Solution

The integrand is a sine times a cosine of a different angle, so we use the product-to-sum identity

\sin a x \cos b x = \frac{1}{2} [\sin (a - b) x + \sin (a + b) x] .

Here a is the coefficient of $x$ inside the sine and b is the coefficient of $x$ inside the cosine. Our integrand is $\sin 4 x \cos 2 x$ , so the sine carries $4 x$ and the cosine carries $2 x$ , which means

a = 4, b = 2, a - b = 2, a + b = 6.

Substituting these into the identity gives $\sin 4 x \cos 2 x = \frac{1}{2} [\sin 2 x + \sin 6 x]$ , and therefore

\begin{aligned} \int\sin 4x\cos 2x\,dx &= \frac{1}{2}\int\left[\sin 2x+\sin 6x\right]\,dx\\ &= \frac{1}{2}\left(-\frac{\cos 2x}{2}-\frac{\cos 6x}{6}\right)+C\\ &= -\frac{\cos 2x}{4}-\frac{\cos 6x}{12}+C. \end{aligned}

Evaluate $\int \sin x \cos 4 x d x$ . (Here the sine has the smaller coefficient, so watch the sign.)

Solution

We again use the product-to-sum identity for a sine times a cosine,

\sin a x \cos b x = \frac{1}{2} [\sin (a - b) x + \sin (a + b) x],

where a is the coefficient of $x$ inside the sine and b is the coefficient of $x$ inside the cosine. Our integrand is $\sin x \cos 4 x$ . The sine carries $1 \cdot x$ and the cosine carries $4 x$ , so

a = 1, b = 4, a - b = - 3, a + b = 5.

This time $a - b$ is negative. Since the sine is an odd function, $\sin (- 3 x) = - \sin 3 x$ , and the identity becomes

\sin x \cos 4 x = \frac{1}{2} [\sin (- 3 x) + \sin 5 x] = \frac{1}{2} [\sin 5 x - \sin 3 x] .

Therefore

\begin{aligned} \int\sin x\cos 4x\,dx &= \frac{1}{2}\int\left[\sin 5x-\sin 3x\right]\,dx\\ &= \frac{1}{2}\left(-\frac{\cos 5x}{5}+\frac{\cos 3x}{3}\right)+C\\ &= \frac{\cos 3x}{6}-\frac{\cos 5x}{10}+C. \end{aligned}

Evaluate $\int \sin 2 x \sin 5 x d x$ .

Solution

The integrand is a product of two sines, so we use the product-to-sum identity

\sin a x \sin b x = \frac{1}{2} [\cos (a - b) x - \cos (a + b) x] .

Here a is the coefficient of $x$ in the first sine and b is the coefficient of $x$ in the second sine. Our integrand is $\sin 2 x \sin 5 x$ , so

a = 2, b = 5, a - b = - 3, a + b = 7.

Because the cosine is an even function, $\cos (- 3 x) = \cos 3 x$ , so the negative difference causes no trouble here. The identity gives $\sin 2 x \sin 5 x = \frac{1}{2} [\cos 3 x - \cos 7 x]$ , and therefore

\begin{aligned} \int\sin 2x\sin 5x\,dx &= \frac{1}{2}\int\left[\cos 3x-\cos 7x\right]\,dx\\ &= \frac{1}{2}\left(\frac{\sin 3x}{3}-\frac{\sin 7x}{7}\right)+C\\ &= \frac{\sin 3x}{6}-\frac{\sin 7x}{14}+C. \end{aligned}

Evaluate $\int \sin 7 x \sin 3 x d x$ .

Solution

Two sines again, so we use

\sin a x \sin b x = \frac{1}{2} [\cos (a - b) x - \cos (a + b) x],

with a the coefficient of $x$ in the first sine and b the coefficient of $x$ in the second. Our integrand is $\sin 7 x \sin 3 x$ , so

a = 7, b = 3, a - b = 4, a + b = 10.

The identity gives $\sin 7 x \sin 3 x = \frac{1}{2} [\cos 4 x - \cos 10 x]$ . Note the minus sign between the two cosines; this is what distinguishes the sine-times-sine identity from the cosine-times-cosine one. Therefore

\begin{aligned} \int\sin 7x\sin 3x\,dx &= \frac{1}{2}\int\left[\cos 4x-\cos 10x\right]\,dx\\ &= \frac{1}{2}\left(\frac{\sin 4x}{4}-\frac{\sin 10x}{10}\right)+C\\ &= \frac{\sin 4x}{8}-\frac{\sin 10x}{20}+C. \end{aligned}

Evaluate $\int \cos 5 x \cos x d x$ .

Solution

The integrand is a product of two cosines, so we use the product-to-sum identity

\cos a x \cos b x = \frac{1}{2} [\cos (a - b) x + \cos (a + b) x] .

Here a is the coefficient of $x$ in the first cosine and b is the coefficient of $x$ in the second cosine. Our integrand is $\cos 5 x \cos x$ , so the first cosine carries $5 x$ and the second carries $1 \cdot x$ , which means

a = 5, b = 1, a - b = 4, a + b = 6.

The identity gives $\cos 5 x \cos x = \frac{1}{2} [\cos 4 x + \cos 6 x]$ , and therefore

\begin{aligned} \int\cos 5x\cos x\,dx &= \frac{1}{2}\int\left[\cos 4x+\cos 6x\right]\,dx\\ &= \frac{1}{2}\left(\frac{\sin 4x}{4}+\frac{\sin 6x}{6}\right)+C\\ &= \frac{\sin 4x}{8}+\frac{\sin 6x}{12}+C. \end{aligned}

Evaluate $\int \cos 3 x \cos 4 x d x$ .

Solution

Two cosines, so we use

\cos a x \cos b x = \frac{1}{2} [\cos (a - b) x + \cos (a + b) x],

with a the coefficient of $x$ in the first cosine and b the coefficient of $x$ in the second. Our integrand is $\cos 3 x \cos 4 x$ , so

a = 3, b = 4, a - b = - 1, a + b = 7.

Since the cosine is even, $\cos (- x) = \cos x$ , so the identity gives $\cos 3 x \cos 4 x = \frac{1}{2} [\cos x + \cos 7 x]$ . Therefore

\begin{aligned} \int\cos 3x\cos 4x\,dx &= \frac{1}{2}\int\left[\cos x+\cos 7x\right]\,dx\\ &= \frac{1}{2}\left(\sin x+\frac{\sin 7x}{7}\right)+C\\ &= \frac{\sin x}{2}+\frac{\sin 7x}{14}+C. \end{aligned}

Evaluate $\int \sin \frac{x}{2} \cos \frac{3 x}{2} d x$ . (The coefficients need not be whole numbers.)

Solution

The integrand is a sine times a cosine, so we use

\sin a x \cos b x = \frac{1}{2} [\sin (a - b) x + \sin (a + b) x],

where a is the coefficient of $x$ inside the sine and b is the coefficient of $x$ inside the cosine. Our integrand is $\sin \frac{x}{2} \cos \frac{3 x}{2}$ . Writing $\frac{x}{2} = \frac{1}{2} \cdot x$ and $\frac{3 x}{2} = \frac{3}{2} \cdot x$ , we read off

a = \frac{1}{2}, b = \frac{3}{2}, a - b = - 1, a + b = 2.

Nothing in the identity requires a and b to be integers. Using $\sin (- x) = - \sin x$ , the identity gives

\sin \frac{x}{2} \cos \frac{3 x}{2} = \frac{1}{2} [\sin (- x) + \sin 2 x] = \frac{1}{2} [\sin 2 x - \sin x] .

Therefore

\begin{aligned} \int\sin\frac{x}{2}\cos\frac{3x}{2}\,dx &= \frac{1}{2}\int\left[\sin 2x-\sin x\right]\,dx\\ &= \frac{1}{2}\left(-\frac{\cos 2x}{2}+\cos x\right)+C\\ &= \frac{\cos x}{2}-\frac{\cos 2x}{4}+C. \end{aligned}

Apply the product-to-sum identity to $\int \cos 3 x \cos 3 x d x$ , and check that it agrees with the half-angle method.

Solution

The integrand is a product of two cosines, so the relevant identity is

\cos a x \cos b x = \frac{1}{2} [\cos (a - b) x + \cos (a + b) x],

with a the coefficient of $x$ in the first cosine and b the coefficient of $x$ in the second. Both factors are $\cos 3 x$ , so this time the two coefficients are equal:

a = 3, b = 3, a - b = 0, a + b = 6.

Since $\cos (0 \cdot x) = \cos 0 = 1$ , the identity collapses to $\cos 3 x \cos 3 x = \frac{1}{2} [1 + \cos 6 x]$ , and therefore

\begin{aligned} \int\cos^{2}3x\,dx &= \frac{1}{2}\int\left[1+\cos 6x\right]\,dx\\ &= \frac{x}{2}+\frac{\sin 6x}{12}+C. \end{aligned}

Check against the half-angle method. The half-angle identity applied to the angle $3 x$ , whose double is $6 x$ , reads

\cos^{2} 3 x = \frac{1 + \cos 6 x}{2},

which is the same integrand, so it gives the same answer. This confirms that the half-angle identities are exactly the special case $a = b$ of the product-to-sum identities.

In an alternating-current circuit, the instantaneous power is the product of a voltage $\cos ω t$ and a current $\cos 2 ω t$ . Find $\int \cos ω t \cos 2 ω t d t$ , where $ω > 0$ is a constant.

Solution

The variable of integration is $t$ , and the integrand is a product of two cosines, so we use the product-to-sum identity in the form

\cos A \cos B = \frac{1}{2} [\cos (A - B) + \cos (A + B)],

where A is the whole angle inside the first cosine and B is the whole angle inside the second. Our integrand is $\cos ω t \cos 2 ω t$ , so

A = ω t, B = 2 ω t, A - B = - ω t, A + B = 3 ω t .

Since the cosine is even, $\cos (- ω t) = \cos ω t$ , and the identity gives

\cos ω t \cos 2 ω t = \frac{1}{2} [\cos ω t + \cos 3 ω t] .

Therefore

\begin{aligned} \int\cos\omega t\cos 2\omega t\,dt &= \frac{1}{2}\int\left[\cos\omega t+\cos 3\omega t\right]\,dt\\ &= \frac{1}{2}\left(\frac{\sin\omega t}{\omega}+\frac{\sin 3\omega t}{3\omega}\right)+C\\ &= \frac{\sin\omega t}{2\omega}+\frac{\sin 3\omega t}{6\omega}+C. \end{aligned}

Evaluate $\int \tan^{2} x d x$ .

Solution

There is no odd power of $\tan x$ and no factor of $\sec x$ at all, so neither case applies directly. Instead we convert the whole integrand into something we can integrate, using the Pythagorean identity

\sec^{2} x = 1 + \tan^{2} x, that is, \tan^{2} x = \sec^{2} x - 1.

Therefore

\begin{aligned} \int\tan^{2}x\,dx &= \int\left(\sec^{2}x-1\right)\,dx\\ &= \tan x-x+C. \end{aligned}

Evaluate $\int \cot^{2} x d x$ .

Solution

Here we use the companion Pythagorean identity, the one relating the cotangent and the cosecant:

\csc^{2} x = 1 + \cot^{2} x, that is, \cot^{2} x = \csc^{2} x - 1.

Therefore

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

Evaluate $\int \tan^{3} x d x$ .

Solution

Split off a factor of $\tan^{2} x$ and replace it using the Pythagorean identity

\sec^{2} x = 1 + \tan^{2} x, that is, \tan^{2} x = \sec^{2} x - 1.

Then

\begin{aligned} \int\tan^{3}x\,dx &= \int\tan x\left(\sec^{2}x-1\right)\,dx\\ &= \int\tan x\sec^{2}x\,dx-\int\tan x\,dx. \end{aligned}

For the first integral, let $u = \tan x$ , so $d u = \sec^{2} x d x$ and $\int u d u = \frac{u^{2}}{2}$ . For the second, we use the formula established earlier in this section. Therefore

\int \tan^{3} x d x = \frac{\tan^{2} x}{2} - \ln | \sec x | + C .

Evaluate $\int \sec^{4} x d x$ .

Solution

The exponent of $\sec x$ is even, so Case 1 applies even though there is no factor of $\tan x$ (that is, $m = 0$ ). We peel off one $\sec^{2} x$ for the differential and convert the remaining $\sec^{2} x$ into tangents with the Pythagorean identity

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

Therefore

\int \sec^{4} x d x = \int (1 + \tan^{2} x) \sec^{2} x d x .

Let $u = \tan x$ , so $d u = \sec^{2} x d x$ :

\begin{aligned} \int(1+u^{2})\,du &= u+\frac{u^{3}}{3}+C\\ &= \tan x+\frac{\tan^{3}x}{3}+C. \end{aligned}

Evaluate $\int \tan^{4} x \sec^{4} x d x$ .

Solution

The exponent of $\sec x$ is even, so we use Case 1: peel off one $\sec^{2} x$ for the differential, and convert the remaining $\sec^{2} x$ into tangents with the Pythagorean identity

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

Therefore

\begin{aligned} \int\tan^{4}x\sec^{4}x\,dx &= \int\tan^{4}x\,(1+\tan^{2}x)\sec^{2}x\,dx. \end{aligned}

With $u = \tan x$ and $d u = \sec^{2} x d x$ ,

\begin{aligned} \int u^{4}(1+u^{2})\,du &= \int\left(u^{4}+u^{6}\right)\,du\\ &= \frac{u^{5}}{5}+\frac{u^{7}}{7}+C\\ &= \frac{\tan^{5}x}{5}+\frac{\tan^{7}x}{7}+C. \end{aligned}

Evaluate $\int \tan^{5} x \sec x d x$ .

Solution

The exponent of $\tan x$ is odd, so we use Case 2: peel off a factor of $\sec x \tan x$ for the differential, and convert the remaining $\tan^{4} x$ into secants with the Pythagorean identity

\sec^{2} x = 1 + \tan^{2} x, that is, \tan^{2} x = \sec^{2} x - 1,

which squared gives $\tan^{4} x = {(\tan^{2} x)}^{2} = (\sec^{2} x - 1)^{2}$ . Therefore

\begin{aligned} \int\tan^{5}x\sec x\,dx &= \int\tan^{4}x\cdot\sec x\tan x\,dx\\ &= \int(\sec^{2}x-1)^{2}\sec x\tan x\,dx. \end{aligned}

Let $u = \sec x$ , so $d u = \sec x \tan x d x$ :

\begin{aligned} \int(u^{2}-1)^{2}\,du &= \int\left(u^{4}-2u^{2}+1\right)\,du\\ &= \frac{u^{5}}{5}-\frac{2u^{3}}{3}+u+C\\ &= \frac{\sec^{5}x}{5}-\frac{2\sec^{3}x}{3}+\sec x+C. \end{aligned}

Evaluate $\int e^{x} \tan (e^{x}) d x$ .

Solution

Let $u = e^{x}$ , so $d u = e^{x} d x$ . Then

\begin{aligned} \int e^{x}\tan(e^{x})\,dx &= \int\tan u\,du\\ &= \ln|\sec u|+C\\ &= \ln\left|\sec(e^{x})\right|+C. \end{aligned}

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

Solution

The numerator is exactly the derivative of the denominator. Let $u = 1 + \tan x$ , so $d u = \sec^{2} x d x$ :

\begin{aligned} \int\frac{\sec^{2}x}{1+\tan x}\,dx &= \int\frac{du}{u}\\ &= \ln|u|+C\\ &= \ln|1+\tan x|+C. \end{aligned}

Evaluate $\int \frac{\sin^{3} x}{\cos^{4} x} d x$ .

Solution

The exponent of $\sin x$ is odd, so we peel off one factor of $\sin x$ for the differential and convert the remaining $\sin^{2} x$ into cosines with the Pythagorean identity

\sin^{2} x + \cos^{2} x = 1, that is, \sin^{2} x = 1 - \cos^{2} x .

Therefore

\int \frac{\sin^{3} x}{\cos^{4} x} d x = \int \frac{1 - \cos^{2} x}{\cos^{4} x} \sin x d x .

Let $u = \cos x$ , so $d u = - \sin x d x$ :

\begin{aligned} \int\frac{1-u^{2}}{u^{4}}(-du) &= \int\left(u^{-2}-u^{-4}\right)\,du\\ &= -u^{-1}+\frac{u^{-3}}{3}+C\\ &= -\frac{1}{\cos x}+\frac{1}{3\cos^{3}x}+C\\ &= -\sec x+\frac{\sec^{3}x}{3}+C. \end{aligned}

The same integral can be written as $\int \tan^{3} x \sec x d x$ , and Case 2 gives the same answer.

Evaluate $\int \sin 4 x \cos 4 x d x$ in two different ways and reconcile the answers.

Solution

Method (a): substitution. Let $u = \sin 4 x$ , so $d u = 4 \cos 4 x d x$ :

\begin{aligned} \int\sin 4x\cos 4x\,dx &= \frac{1}{4}\int u\,du\\ &= \frac{u^{2}}{8}+C\\ &= \frac{\sin^{2}4x}{8}+C. \end{aligned}

Method (b): double-angle identity. We use

\sin 2 θ = 2 \sin θ \cos θ, that is, \sin θ \cos θ = \frac{1}{2} \sin 2 θ,

where $θ$ is the angle shared by the sine and the cosine. Here $θ = 4 x$ , so $2 θ = 8 x$ and $\sin 4 x \cos 4 x = \frac{1}{2} \sin 8 x$ . Therefore

\begin{aligned} \int\sin 4x\cos 4x\,dx &= \frac{1}{2}\int\sin 8x\,dx\\ &= -\frac{\cos 8x}{16}+C_{1}. \end{aligned}

Reconciling. We use the double-angle identity for the cosine in the form

\cos 2 θ = 1 - 2 \sin^{2} θ,

where $θ$ is the angle inside the sine on the right. Taking $θ = 4 x$ gives $2 θ = 8 x$ , so $\cos 8 x = 1 - 2 \sin^{2} 4 x$ and

- \frac{\cos 8 x}{16} = - \frac{1 - 2 \sin^{2} 4 x}{16} = \frac{\sin^{2} 4 x}{8} - \frac{1}{16},

so the two answers differ only by the constant $\frac{1}{16}$ , which is absorbed into the constant of integration.

Integrals That Follow Directly From Derivatives

The Integrals of tan x and cot x

The Integral of sec x tan x

The Integrals of sec x and csc x

The Integrals of sin²x and cos²x

Products of Sines and Cosines of Different Angles

Integrating sin^m x cosⁿ x When Either m or n Is a Positive Odd Integer

Integrating sin^m x cosⁿ x When Both m and n Are Positive Even Integers

Integrating tan^m x secⁿ x When m Is a Positive Odd Integer or n Is a Positive Even Integer

Exercises

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Integrals Of Trigonometric Functions

Integrals That Follow Directly From Derivatives

The Integrals of tan x and cot x

The Integral of sec x tan x

The Integrals of sec x and csc x

The Integrals of sin²x and cos²x

Products of Sines and Cosines of Different Angles

Integrating sinm x cosn x When Either m or n Is a Positive Odd Integer

Integrating sinm x cosn x When Both m and n Are Positive Even Integers

Integrating tanm x secn x When m Is a Positive Odd Integer or n Is a Positive Even Integer

Exercises

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Integrating sin^m x cosⁿ x When Either m or n Is a Positive Odd Integer

Integrating sin^m x cosⁿ x When Both m and n Are Positive Even Integers

Integrating tan^m x secⁿ x When m Is a Positive Odd Integer or n Is a Positive Even Integer