Basic Integration Formulas

Integration is an indirect process: to integrate, we must answer the question "What function, when differentiated, yields the given integrand?" This section presents the essential formulas for the most common elementary functions.

Formula	Name
$\int x^{α} d x = \frac{x^{α + 1}}{α + 1} + C$	Power Rule ( $α \neq - 1$ )
$\int \frac{d x}{x} = \ln \| x \| + C$	Logarithm Rule
$\int \cos x d x = \sin x + C$	Cosine Integral
$\int \sin x d x = - \cos x + C$	Sine Integral
$\int \sec^{2} x d x = \tan x + C$	Secant Squared
$\int \csc^{2} x d x = - \cot x + C$	Cosecant Squared
$\int \frac{d x}{1 + x^{2}} = \arctan x + C$	Arctangent
$\int \frac{d x}{\sqrt{1 - x^{2}}} = \arcsin x + C$	Arcsine
$\int e^{x} d x = e^{x} + C$	Exponential

Integration Is an Indirect Process

Differentiation is a direct process. To differentiate, we need to evaluate

lim_{Δ x \to 0} \frac{f (x + Δ x) - f (x)}{Δ x} .

In contrast, integration is an indirect process with no corresponding general rule. To integrate, we have to answer the question: What function when differentiated will yield the given integrand? So integration is actually a process of finding our way back. This is often difficult and cannot always be solved; in fact, there are many comparatively simple functions whose integrals cannot be expressed in terms of elementary functions.¹

¹Elementary functions are the class of functions that can be constructed in a finite number of steps from the basic algebraic operations (addition, subtraction, multiplication, division, and taking roots), together with the exponential function, the natural logarithm, the trigonometric functions, and their inverses. Every polynomial is elementary. So is $e^{- x^{2}}$ , or $\ln (\sin x + \sqrt{1 + x^{2}})$ , no matter how deeply nested.

To integrate a given function, we need to know the integrals of a few basic functions (obtained through guesswork or otherwise). In this section, we present the integrals of these basic functions that are necessary for calculus. We then attempt to reduce a given integral to one of these familiar integrals by using integral properties and integration techniques, some of which will be discussed in this chapter.

Integrals of Basic Functions

In this section we study the integrals of the most common elementary functions. These formulas serve as building blocks for evaluating a wide range of integrals.

In fact, with these formulas in hand, one can evaluate the vast majority of integrals encountered in a standard calculus course. When an integral does not match one of these forms directly, our strategy is to reduce it to one that does, using the properties and techniques discussed in the sections and chapters that follow.

The Power Rule for Integration

What is $\int x^{2} d x$ ? Recall that

\frac{d}{d x} (\frac{1}{3} x^{3}) = x^{2},

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

The same idea works for any real exponent. For $r \in ℝ$ , $r \neq 0$ , we have

d (x^{r}) = r x^{r - 1} d x,

or equivalently,

d (\frac{x^{r}}{r}) = x^{r - 1} d x,

which gives

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

Setting $r - 1 = α$ yields the Power Rule for Integration:

\begin{aligned} \int x^\alpha\,dx &= \frac{x^{\alpha+1}}{\alpha+1} + C, \\ &\qquad \alpha \in \mathbb{R},\ \alpha \neq -1. \end{aligned}

The procedure is easy to remember: increase the exponent by one, then divide by that new exponent.

The case $α = - 1$ . The power rule breaks down when $α = - 1$ , because $α + 1 = 0$ . We handle this case separately.

For $x > 0$ , we have $\frac{d}{d x} \ln x = \frac{1}{x}$ , so

\int \frac{d x}{x} = \ln x + C (x > 0) .

For $x < 0$ , the quantity $- x$ is positive, so $\ln (- x)$ is defined. Differentiating gives

\frac{d}{d x} \ln (- x) = \frac{- 1}{- x} = \frac{1}{x},

and therefore

\int \frac{d x}{x} = \ln (- x) + C (x < 0) .

The two cases combine into a single formula:

\int \frac{d x}{x} = \ln | x | + C,

valid on any interval that does not contain $x = 0$ .

The integral $\int \frac{1}{x} d x$ is often written as $\int \frac{d x}{x}$ . The same shorthand is used whenever the integrand is a fraction.

Collecting both subcases, we have the complete power rule:

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{ \int x^\alpha\,dx = \begin{cases} \dfrac{x^{\alpha+1}}{\alpha+1}+C, & \alpha \neq -1 \\[6pt] \ln|x|+C, & \alpha = -1 \end{cases}}

First Examples

Find $\int \sqrt[3]{x} d x$ .

Solution

\begin{aligned} \int\sqrt[3]{x}\,dx &= \int x^{1/3}\,dx \\ &= \frac{1}{\tfrac{1}{3}+1}\,x^{1/3+1}+C \\ &= \frac{3}{4}\,x^{4/3}+C. \end{aligned}

Verification. Differentiating the result gives back the integrand:

\begin{aligned} \frac{d}{dx}\!\left(\frac{3}{4}x^{4/3}+C\right) &= \frac{3}{4}\cdot\frac{4}{3}\,x^{1/3} \\ &= x^{1/3} = \sqrt[3]{x}. \end{aligned}

The integrand is recovered. ✓

Find $\int \frac{d θ}{\sqrt{θ}}$ .

Solution

\begin{aligned} \int\frac{1}{\sqrt{\theta}}\,d\theta &= \int\theta^{-1/2}\,d\theta \\ &= \frac{1}{-\tfrac{1}{2}+1}\,\theta^{-\frac{1}{2}+1}+C \\ &= 2\,\theta^{1/2}+C \\ &= 2\sqrt{\theta}+C. \end{aligned}

To verify, differentiate the result and confirm you recover the integrand.

Trigonometric and Inverse Trigonometric Functions

Reading the differentiation formulas for $\sin x$ , $\cos x$ , $\arctan x$ , and $\arcsin x$ from right to left gives the following integrals immediately.

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

Since $\frac{d}{d x} \arctan x = \frac{1}{1 + x^{2}}$ , we have

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

Since $\frac{d}{d x} \arcsin x = \frac{1}{\sqrt{1 - x^{2}}}$ , we have

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

Also since $\frac{d}{d x} \arccos x = - \frac{1}{\sqrt{1 - x^{2}}}$ , or equivalently $\frac{d}{d x} (- \arccos x) = \frac{1}{\sqrt{1 - x^{2}}}$ , we can write

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

Recall that $\arctan$ , $\arcsin$ , and $\arccos$ are sometimes written $\tan^{- 1}$ , $\sin^{- 1}$ , $\cos^{- 1}$ respectively.

The Exponential Function

Because $e^{x}$ is its own derivative, it is also its own antiderivative:

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

In each formula above, when the constant $C$ is omitted, the right-hand side should be understood as one particular antiderivative of the integrand, not the most general one.

Properties of Indefinite Integrals

So far each integrand involved a single term. Most integrands, however, consist of sums or constant multiples of simpler functions. The following theorem makes it possible to break such integrals apart and handle each piece separately.

The following properties of antiderivatives are direct consequences of those of derivatives.

If $f (x)$ and $g (x)$ have indefinite integrals in the same interval $I$ , and if $k$ is a constant, then

\int kf(x)\,dx = k\int f(x)\,dx. \tag{a}\begin{aligned} \int[f(x)+g(x)]\,dx &= \int f(x)\,dx \\ &\quad + \int g(x)\,dx. \end{aligned} \tag{b}

Proof

To prove the above relationships, we show the integral functions in the left-hand side (LHS) and the right-hand side (RHS) have the same derivatives, and for this we use the properties of differentiation and the definition $\int f (x) d x = F (x) + C$ .

(a)

\begin{aligned} \text{derivative of LHS} &= \frac{d}{dx}\left(\int kf(x)\,dx\right) \\ &= kf(x) \end{aligned}\begin{aligned} \text{derivative of RHS} &= \frac{d}{dx}\left(k\int f(x)\,dx\right) \\ &= \left(\frac{dk}{dx}\right)\int f(x)\,dx \\ &\quad + k\left(\frac{d}{dx}\int f(x)\,dx\right) \\ &= 0 + kf(x). \end{aligned}

Since the derivatives are the same, we conclude that $\int k f (x) d x = k \int f (x) d x .$

(b)

\begin{aligned} \text{derivative of LHS} &= \frac{d}{dx}\int[f(x)+g(x)]\,dx \\ &= f(x)+g(x) \end{aligned}\begin{aligned} \text{derivative of RHS} &= \frac{d}{dx}\left(\int f(x)\,dx + \int g(x)\,dx\right) \\ &= \frac{d}{dx}\int f(x)\,dx + \frac{d}{dx}\int g(x)\,dx \\ &= f(x)+g(x). \end{aligned}

For the second line, we have used the Sum Rule. Since the derivatives of the left-hand side and of the right-hand side are the same, we conclude

\begin{aligned} \int[f(x)+g(x)]\,dx &= \int f(x)\,dx \\ &\quad + \int g(x)\,dx. \end{aligned}

The first property means that the integral of a constant times a function is equal to the constant times the integral of that function. In other words, this tells us that we can pull a constant out of an integral. Notice that this does not hold true if $k$ is a variable.
The second property means that the integral of the sum or difference of two functions is equal to the sum or difference of their respective integrals.
In the above theorem, it is assumed that the arbitrary constants are adjusted properly. Thus, formula (b) asserts that the sum of any antiderivative of $f (x)$ and any antiderivative of $g (x)$ is an antiderivative of $f (x) + g (x)$ .

We can combine formulas (a) and (b) and write down for any finite number of functions:

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{\begin{aligned} \int[a_1 f_1(x) &+ a_2 f_2(x) \\ &\quad + \cdots + a_n f_n(x)]\,dx \\ &= a_1\int f_1(x)\,dx + a_2\int f_2(x)\,dx \\ &\quad + \cdots + a_n\int f_n(x)\,dx. \end{aligned}} \tag{c}

In particular, the power rule for integration together with the linearity of the integral gives the integral of any polynomial at once:

\begin{aligned} \int\!\bigl(a_0 x^n &+ a_1 x^{n-1} + \cdots \\ &\quad + a_{n-1}x + a_n\bigr)\,dx \\ &= \frac{a_0}{n+1}x^{n+1} + \frac{a_1}{n}x^n \\ &\quad + \cdots + \frac{a_{n-1}}{2}x^2 + a_n x + C. \end{aligned}

Each term contributes its own constant of integration, but all of them merge into the single constant $C$ .

Further Examples

The examples that follow all require the linearity properties above, combined with the basic formulas from the table. The recurring strategy is the same: rewrite the integrand so that each term matches one of the known forms, then integrate term by term.

Find $\int (\frac{2 a}{\sqrt{y}} - \frac{b}{y} + c \sqrt[3]{y^{2}}) d y$ , where $a$ , $b$ , and $c$ are constants.

Solution

\begin{aligned} \int\!\left(\frac{2a}{\sqrt{y}}-\frac{b}{y}+c\sqrt[3]{y^2}\right)\,dy &= 2a\int y^{-1/2}\,dy - b\int\frac{dy}{y} \\ &\quad + c\int y^{2/3}\,dy \\ &= 2a\cdot 2\,y^{1/2} - b\ln|y| \\ &\quad + c\cdot\tfrac{3}{5}\,y^{5/3}+C \\ &= 4a\sqrt{y} - b\ln|y| \\ &\quad + \tfrac{3c}{5}\,\sqrt[3]{y^5}+C. \end{aligned}

Find $\int (5 \cos t + \frac{3 t^{2} + \sqrt{t}}{t}) d t$ .

Solution

Splitting the fraction in the second term gives $\frac{3 t^{2} + \sqrt{t}}{t} = 3 t + t^{- 1 / 2}$ , so

\begin{aligned} \int\!\left(5\cos t + \frac{3t^2+\sqrt{t}}{t}\right)\,dt &= 5\int\cos t\,dt + 3\int t\,dt \\ &\quad + \int t^{-1/2}\,dt \\ &= 5\sin t + \tfrac{3}{2}t^2 + 2\sqrt{t} + C. \end{aligned}

Find $\int [5 (1 + u^{2})^{- 1} - e^{u}] d u$ .

Solution

\begin{aligned} \int\!\left[\frac{5}{1+u^2} - e^u\right]\,du &= 5\int\frac{du}{1+u^2} - \int e^u\,du \\ &= 5\arctan u - e^u + C. \end{aligned}

Find $\int (\frac{x^{2} + 3}{x} - \frac{2}{\sqrt{1 - x^{2}}}) d x$ .

Solution

Writing $\frac{x^{2} + 3}{x} = x + \frac{3}{x}$ , we get

\begin{aligned} \int\!\left(\frac{x^2+3}{x} - \frac{2}{\sqrt{1-x^2}}\right)\,dx &= \int\!\left(x + \frac{3}{x}\right)\,dx \\ &\quad - 2\int\frac{dx}{\sqrt{1-x^2}} \\ &= \frac{1}{2}x^2 + 3\ln|x| \\ &\quad - 2\arcsin x + C. \end{aligned}

Sometimes the integrand does not match any entry in the table directly, but a simple algebraic manipulation reveals a combination that does. The next two examples illustrate this.

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

Solution

Write $x^{2} = (x^{2} + 1) - 1$ in the numerator:

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

Find $\int \frac{x^{4}}{x^{2} + 1} d x$ .

Solution

Write $x^{4} = (x^{4} - 1) + 1 = (x^{2} - 1) (x^{2} + 1) + 1$ , so that

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

The final example shows that a trigonometric identity can serve the same purpose as an algebraic manipulation: it rewrites the integrand in a form the table can handle.

Find $\int \sqrt{1 - \cos 2 x} d x$ for $0 \leq x \leq π$ .

Solution

The half-angle identity $\sin^{2} x = \frac{1}{2} (1 - \cos 2 x)$ gives $1 - \cos 2 x = 2 \sin^{2} x$ , so

\begin{aligned} \int\sqrt{1-\cos 2x}\;dx &= \sqrt{2}\int|\sin x|\,dx. \end{aligned}

Since $\sin x \geq 0$ on $[0, π]$ , we have $| \sin x | = \sin x$ there, and therefore

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

Table of Integration Formulas

Exercises

Evaluate each integral using the Power Rule for Integration:

(a) $\int x^{7} d x$

(b) $\int \frac{d x}{x^{5}}$

(c) $\int \sqrt[5]{x} d x$

(d) $\int \frac{d x}{\sqrt[3]{x^{2}}}$

(e) $\int x^{\sqrt{2}} d x$

(f) $\int \frac{d x}{x^{2} \sqrt{x}}$

Answer

(a) $\frac{x^{8}}{8} + C$

(b) $- \frac{1}{4 x^{4}} + C$

(c) $\frac{5}{6} x^{6 / 5} + C$

(d) $3 \sqrt[3]{x} + C$

(e) $(\sqrt{2} - 1) x^{\sqrt{2} + 1} + C$

(f) $- \frac{2}{3 x \sqrt{x}} + C$

Solution

In every part, rewrite the integrand as a single power $x^{α}$ , then apply $\int x^{α} d x = \frac{x^{α + 1}}{α + 1} + C$ : increase the exponent by one, then divide by the new exponent.

(a)

\int x^{7} d x = \frac{x^{7 + 1}}{7 + 1} + C = \frac{x^{8}}{8} + C .

(b) Here

\frac{1}{x^{5}} = x^{- 5}

, so

α = - 5

and

α + 1 = - 4 \neq 0

\int x^{- 5} d x = \frac{x^{- 4}}{- 4} + C = - \frac{1}{4 x^{4}} + C .

(c) Here

\sqrt[5]{x} = x^{1 / 5}

: \begin{aligned} \int x^{1/5}\,dx &= \frac{x^{6/5}}{\tfrac{1}{5}+1}+C \\ &= \frac{5}{6}x^{6/5}+C. \end{aligned}
(d) Here

\frac{1}{\sqrt[3]{x^{2}}} = x^{- 2 / 3}

: \begin{aligned} \int x^{-2/3}\,dx &= \frac{x^{1/3}}{-\tfrac{2}{3}+1}+C \\ &= 3x^{1/3}+C = 3\sqrt[3]{x}+C. \end{aligned}
(e) The exponent may be any real number, rational or not: \begin{aligned} \int x^{\sqrt{2}}\,dx &= \frac{x^{\sqrt{2}+1}}{\sqrt{2}+1}+C \\ &= (\sqrt{2}-1)\,x^{\sqrt{2}+1}+C, \end{aligned}

where we rationalized using $\frac{1}{\sqrt{2} + 1} = \sqrt{2} - 1$ .

(f) Combine the powers first:

\frac{1}{x^{2} \sqrt{x}} = \frac{1}{x^{2} \cdot x^{1 / 2}} = x^{- 5 / 2}

. Then \begin{aligned} \int x^{-5/2}\,dx &= \frac{x^{-3/2}}{-\tfrac{3}{2}}+C \\ &= -\frac{2}{3}x^{-3/2}+C \\ &= -\frac{2}{3x\sqrt{x}}+C. \end{aligned}

The power rule fails for $α = - 1$ . Evaluate the following, and state in one sentence why the case $α = - 1$ has to be treated separately:

(a) $\int \frac{2}{x} d x$

(b) $\int \frac{x^{2} + 3 x + 1}{x} d x$

(c) $\int (\frac{1}{x} - \frac{1}{x^{2}}) d x$

Answer

(a) $2 \ln | x | + C$

(b) $\frac{x^{2}}{2} + 3 x + \ln | x | + C$

(c) $\ln | x | + \frac{1}{x} + C$

Reason: for $α = - 1$ the new exponent is $α + 1 = 0$ , so the formula would require dividing by zero.

Solution

Why $α = - 1$ is special. The power rule produces $\frac{x^{α + 1}}{α + 1}$ . When $α = - 1$ this is $\frac{x^{0}}{0}$ , which is undefined. The missing antiderivative is supplied by the logarithm: $\frac{d}{d x} \ln | x | = \frac{1}{x}$ for every $x \neq 0$ .

(a) Pull the constant out by property (a) of linearity:

\int \frac{2}{x} d x = 2 \int \frac{d x}{x} = 2 \ln | x | + C .

(b) Split the fraction term by term before integrating:

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

so that

\begin{aligned} \int\frac{x^{2}+3x+1}{x}\,dx &= \int x\,dx + 3\int dx \\ &\quad + \int\frac{dx}{x} \\ &= \frac{x^{2}}{2}+3x+\ln|x|+C. \end{aligned}
(c) The first term is the logarithmic case, the second is an ordinary power with

α = - 2

: \begin{aligned} \int\left(\frac{1}{x}-\frac{1}{x^{2}}\right)dx &= \int\frac{dx}{x} - \int x^{-2}\,dx \\ &= \ln|x| - \frac{x^{-1}}{-1}+C \\ &= \ln|x|+\frac{1}{x}+C. \end{aligned}

Integrate each polynomial. Expand or simplify first where necessary:

(a) $\int (4 x^{3} - 6 x^{2} + 2 x - 7) d x$

(b) $\int (2 x + 1)^{2} d x$

(c) $\int (\sqrt{x} + 1) (\sqrt{x} - 1) d x$

Answer

(a) $x^{4} - 2 x^{3} + x^{2} - 7 x + C$

(b) $\frac{4}{3} x^{3} + 2 x^{2} + x + C$

(c) $\frac{x^{2}}{2} - x + C$

Solution

Every polynomial integrates term by term, by Linearity of Integration combined with the power rule.

(a) \begin{aligned} \int\left(4x^{3}-6x^{2}+2x-7\right)dx &= 4\cdot\frac{x^{4}}{4} - 6\cdot\frac{x^{3}}{3} \\ &\quad + 2\cdot\frac{x^{2}}{2} - 7x + C \\ &= x^{4}-2x^{3}+x^{2}-7x+C. \end{aligned}
(b) The integrand is not a basic form as written, so expand first:

(2 x + 1)^{2} = 4 x^{2} + 4 x + 1

. Then

\int (4 x^{2} + 4 x + 1) d x = \frac{4}{3} x^{3} + 2 x^{2} + x + C .

Warning. It is tempting to write $\frac{(2 x + 1)^{3}}{3}$ , but differentiating that gives $(2 x + 1)^{2} \cdot 2$ , which is twice the integrand. The correct handling of such powers comes with substitution.

(c) The product is a difference of squares:

(\sqrt{x} + 1) (\sqrt{x} - 1) = x - 1

. Hence

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

Split each fraction into separate terms, then integrate:

(a) $\int \frac{x^{3} + 2 x - 4}{x^{2}} d x$

(b) $\int \frac{\sqrt{t} + t}{t^{2}} d t$

(c) $\int \frac{(1 + \sqrt{u})^{2}}{\sqrt{u}} d u$

Answer

(a) $\frac{x^{2}}{2} + 2 \ln | x | + \frac{4}{x} + C$

(b) $- \frac{2}{\sqrt{t}} + \ln | t | + C$

(c) $2 \sqrt{u} + 2 u + \frac{2}{3} u^{3 / 2} + C$

Solution

There is no quotient rule for integration. When the denominator is a single term, divide it into each term of the numerator and integrate the resulting sum.

(a)

\frac{x^{3} + 2 x - 4}{x^{2}} = x + \frac{2}{x} - \frac{4}{x^{2}}

, so \begin{aligned} \int \frac{x^{3}+2x-4}{x^{2}}\,dx &= \int x\,dx + 2\int\frac{dx}{x} \\ &\quad - 4\int x^{-2}\,dx \\ &= \frac{x^{2}}{2}+2\ln|x|+\frac{4}{x}+C. \end{aligned}
(b)

\frac{\sqrt{t} + t}{t^{2}} = t^{1 / 2 - 2} + t^{1 - 2} = t^{- 3 / 2} + t^{- 1}

, so \begin{aligned} \int \frac{\sqrt{t}+t}{t^{2}}\,dt &= \int t^{-3/2}\,dt + \int\frac{dt}{t} \\ &= \frac{t^{-1/2}}{-\tfrac{1}{2}} + \ln|t| + C \\ &= -\frac{2}{\sqrt{t}}+\ln|t|+C. \end{aligned}
(c) Expand the square first, then divide: \begin{aligned} \frac{(1+\sqrt{u})^{2}}{\sqrt{u}} &= \frac{1+2\sqrt{u}+u}{\sqrt{u}} \\ &= u^{-1/2}+2+u^{1/2}. \end{aligned}

Therefore

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

Evaluate the following trigonometric integrals:

(a) $\int (3 \sin x - 2 \cos x) d x$

(b) $\int (\sec^{2} θ + \csc^{2} θ) d θ$

(c) $\int \sec x (\sec x + \tan x) d x$

(d) $\int (\csc x \cot x - \csc^{2} x) d x$

Answer

(a) $- 3 \cos x - 2 \sin x + C$

(b) $\tan θ - \cot θ + C$

(c) $\tan x + \sec x + C$

(d) $- \csc x + \cot x + C$

Solution

Each of these is read directly from the table of integrals. Watch the signs: the two integrals whose answers carry a minus sign are $\int \sin x d x = - \cos x + C$ and $\int \csc^{2} x d x = - \cot x + C$ , together with $\int \csc x \cot x d x = - \csc x + C$ .

(a) \begin{aligned} \int (3\sin x-2\cos x)\,dx &= 3\int\sin x\,dx - 2\int\cos x\,dx \\ &= 3(-\cos x) - 2\sin x + C \\ &= -3\cos x-2\sin x+C. \end{aligned}
(b) \begin{aligned} \int \sec^{2}\theta\,d\theta &+ \int\csc^{2}\theta\,d\theta \\ &= \tan\theta+(-\cot\theta)+C \\ &= \tan\theta-\cot\theta+C. \end{aligned}
(c) Multiply out first:

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

. Hence

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

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

Evaluate the following integrals leading to inverse trigonometric functions:

(a) $\int \frac{5}{1 + x^{2}} d x$

(b) $\int (\frac{2}{\sqrt{1 - x^{2}}} - \frac{3}{1 + x^{2}}) d x$

(c) $\int (\frac{1}{1 + x^{2}} + \frac{1}{\sqrt{1 - x^{2}}}) d x$

Answer

(a) $5 \arctan x + C$

(b) $2 \arcsin x - 3 \arctan x + C$

(c) $\arctan x + \arcsin x + C$

Solution

The two forms to recognize are $\int \frac{d x}{1 + x^{2}} = \arctan x + C$ and $\int \frac{d x}{\sqrt{1 - x^{2}}} = \arcsin x + C$ . The distinguishing feature is the square root: no radical gives arctangent, a radical gives arcsine.

(a) \begin{aligned} \int \frac{5}{1+x^{2}}\,dx &= 5\int\frac{dx}{1+x^{2}} \\ &= 5\arctan x+C. \end{aligned}
(b) \begin{aligned} \int\left(\frac{2}{\sqrt{1-x^{2}}}-\frac{3}{1+x^{2}}\right)dx &= 2\int\frac{dx}{\sqrt{1-x^{2}}} \\ &\quad - 3\int\frac{dx}{1+x^{2}} \\ &= 2\arcsin x-3\arctan x+C. \end{aligned}

Because $\int \frac{d x}{\sqrt{1 - x^{2}}} = - \arccos x + C_{1}$ as well, the answer $- 2 \arccos x - 3 \arctan x + C_{1}$ is equally correct; the two differ only by a constant.

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

valid on $(- 1, 1)$ , where both integrands are defined.

Evaluate the following exponential and hyperbolic integrals:

(a) $\int (e^{x} + x^{e}) d x$

(b) $\int (3 e^{x} - \frac{2}{x}) d x$

(c) $\int (\cosh x - 4 \sinh x) d x$

(d) $\int e^{x + 2} d x$

Answer

(a) $e^{x} + \frac{x^{e + 1}}{e + 1} + C$

(b) $3 e^{x} - 2 \ln | x | + C$

(c) $\sinh x - 4 \cosh x + C$

(d) $e^{x + 2} + C$

Solution

(a) The two terms look alike but are completely different. In

e^{x}

the base is constant and the exponent varies, so it is an exponential function and is its own antiderivative. In

x^{e}

the exponent is the constant

e \approx 2.718

, so it is an ordinary power and the power rule applies:

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

(b) \begin{aligned} \int\left(3e^{x}-\frac{2}{x}\right)dx &= 3\int e^{x}\,dx - 2\int\frac{dx}{x} \\ &= 3e^{x}-2\ln|x|+C. \end{aligned}
(c) From the table,

\int \cosh x d x = \sinh x + C

and

\int \sinh x d x = \cosh x + C

. Unlike the circular functions, neither hyperbolic formula carries a minus sign: \begin{aligned} \int\left(\cosh x-4\sinh x\right)dx \\ = \sinh x-4\cosh x+C. \end{aligned}
(d) Use the law of exponents to expose a constant factor:

e^{x + 2} = e^{2} e^{x}

, and

e^{2}

is just a number. Hence \begin{aligned} \int e^{x+2}\,dx &= e^{2}\int e^{x}\,dx \\ &= e^{2}e^{x}+C = e^{x+2}+C. \end{aligned}

Each integrand below becomes a sum of familiar forms after an algebraic rewriting. Evaluate:

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

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

(c) $\int \frac{x^{4} + 1}{x^{2} + 1} d x$

Answer

(a) $x + \arctan x + C$

(b) $- \frac{1}{x} - \arctan x + C$

(c) $\frac{x^{3}}{3} - x + 2 \arctan x + C$

Solution

The recurring trick is to add and subtract the denominator in the numerator, so that the fraction splits into a polynomial part plus a leftover that matches a table entry.

(a) Write

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

: \begin{aligned} \int \frac{x^{2}+2}{x^{2}+1}\,dx &= \int\left(1+\frac{1}{x^{2}+1}\right)dx \\ &= x+\arctan x+C. \end{aligned}
(b) Write

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

in the numerator and divide: \begin{aligned} \frac{1}{x^{2}(1+x^{2})} &= \frac{(1+x^{2})-x^{2}}{x^{2}(1+x^{2})} \\ &= \frac{1}{x^{2}}-\frac{1}{1+x^{2}}. \end{aligned}

Therefore

\begin{aligned} \int \frac{dx}{x^{2}(1+x^{2})} &= \int x^{-2}\,dx - \int\frac{dx}{1+x^{2}} \\ &= -\frac{1}{x}-\arctan x+C. \end{aligned}
(c) Write

x^{4} + 1 = (x^{4} - 1) + 2 = (x^{2} - 1) (x^{2} + 1) + 2

, so

\frac{x^{4} + 1}{x^{2} + 1} = x^{2} - 1 + \frac{2}{x^{2} + 1},

and hence

\int \frac{x^{4} + 1}{x^{2} + 1} d x = \frac{x^{3}}{3} - x + 2 \arctan x + C .

Use a trigonometric identity to reduce each integrand to a form in the table, then integrate:

(a) $\int \tan^{2} x d x$

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

(c) $\int \sin^{2} \frac{x}{2} d x$

(d) $\int \frac{d x}{1 - \cos 2 x}$

Answer

(a) $\tan x - x + C$

(b) $- \cot x - x + C$

(c) $\frac{x}{2} - \frac{1}{2} \sin x + C$

(d) $- \frac{1}{2} \cot x + C$

Solution

None of these integrands appears in the table, but each is one identity away from something that does.

(a) The Pythagorean identity

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

gives

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

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

\cot^{2} x = \csc^{2} x - 1

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

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

turns the square into a first-power cosine: \begin{aligned} \int \sin^{2}\frac{x}{2}\,dx &= \frac{1}{2}\int\left(1-\cos x\right)dx \\ &= \frac{x}{2}-\frac{1}{2}\sin x+C. \end{aligned}
(d) From

1 - \cos 2 x = 2 \sin^{2} x

we get \begin{aligned} \int \frac{dx}{1-\cos 2x} &= \frac{1}{2}\int\frac{dx}{\sin^{2}x} \\ &= \frac{1}{2}\int\csc^{2}x\,dx \\ &= -\frac{1}{2}\cot x+C. \end{aligned}

Simplify each integrand first, then integrate:

(a) $\int \frac{\cos 2 x}{\cos x - \sin x} d x$

(b) $\int \frac{\cos 2 x}{\cos^{2} x \sin^{2} x} d x$

(c) $\int {(\tan x + \cot x)}^{2} d x$

Answer

(a) $\sin x - \cos x + C$

(b) $- \cot x - \tan x + C$

(c) $\tan x - \cot x + C$

Solution

All three use the double-angle identity $\cos 2 x = \cos^{2} x - \sin^{2} x$ or the Pythagorean identities.

(a) Factor the numerator as a difference of squares: \begin{aligned} \frac{\cos 2x}{\cos x-\sin x} &= \frac{(\cos x-\sin x)(\cos x+\sin x)}{\cos x-\sin x} \\ &= \cos x+\sin x, \end{aligned}

valid wherever $\cos x \neq \sin x$ . Hence

\int (\cos x + \sin x) d x = \sin x - \cos x + C .

(b) Split the numerator over the denominator: \begin{aligned} \frac{\cos^{2}x-\sin^{2}x}{\cos^{2}x\,\sin^{2}x} &= \frac{\cos^{2}x}{\cos^{2}x\,\sin^{2}x} - \frac{\sin^{2}x}{\cos^{2}x\,\sin^{2}x} \\ &= \frac{1}{\sin^{2}x}-\frac{1}{\cos^{2}x} \\ &= \csc^{2}x-\sec^{2}x. \end{aligned}

Therefore

\int (\csc^{2} x - \sec^{2} x) d x = - \cot x - \tan x + C .

(c) Expand the square and use

\tan x \cot x = 1

: \begin{aligned} \left(\tan x+\cot x\right)^{2} &= \tan^{2}x+2\tan x\cot x+\cot^{2}x \\ &= \left(\sec^{2}x-1\right)+2 \\ &\quad + \left(\csc^{2}x-1\right) \\ &= \sec^{2}x+\csc^{2}x. \end{aligned}

Hence

\int {(\tan x + \cot x)}^{2} d x = \tan x - \cot x + C .

Following the worked example on $\int \sqrt{1 - \cos 2 x} d x$ , evaluate:

(a) $\int \sqrt{1 + \cos 2 x} d x$ for $- \frac{π}{2} \leq x \leq \frac{π}{2}$

(b) $\int \sqrt{1 - \cos 2 x} d x$ for $π \leq x \leq 2 π$

Answer

(a) $\sqrt{2} \sin x + C$

(b) $\sqrt{2} \cos x + C$

Solution

The point of both parts is that $\sqrt{v^{2}} = | v |$ , not $v$ . The absolute value must be resolved using the given interval before integrating.

(a) The half-angle identity

\cos^{2} x = \frac{1}{2} (1 + \cos 2 x)

gives

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

, so

\int \sqrt{1 + \cos 2 x} d x = \sqrt{2} \int | \cos x | d x .

On $[- \frac{π}{2}, \frac{π}{2}]$ we have $\cos x \geq 0$ , so $| \cos x | = \cos x$ and

\begin{aligned} \int\sqrt{1+\cos 2x}\,dx &= \sqrt{2}\int\cos x\,dx \\ &= \sqrt{2}\,\sin x+C. \end{aligned}
(b) As in the worked example,

1 - \cos 2 x = 2 \sin^{2} x

, so the integral equals

\sqrt{2} \int | \sin x | d x

. This time the interval is

[π, 2 π]

, where

\sin x \leq 0

, so

| \sin x | = - \sin x

and \begin{aligned} \int\sqrt{1-\cos 2x}\,dx &= -\sqrt{2}\int\sin x\,dx \\ &= -\sqrt{2}\left(-\cos x\right)+C \\ &= \sqrt{2}\,\cos x+C. \end{aligned}

Compare with the worked example, whose answer on $[0, π]$ was $- \sqrt{2} \cos x + C$ . The sign is opposite, which shows that the interval genuinely matters.

Each line below contains a mistake. Identify the error and give the correct result:

(a) $\int \frac{d x}{x} = \frac{x^{0}}{0} + C$

(b) $\int x^{2} \cdot x^{3} d x = \frac{x^{3}}{3} \cdot \frac{x^{4}}{4} + C$

(c) $\int x \cos x d x = x \int \cos x d x = x \sin x + C$

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

Answer

(a) The power rule excludes $α = - 1$ . Correct: $\ln | x | + C$ .

(b) The integral of a product is not the product of the integrals. Correct: $\frac{x^{6}}{6} + C$ .

(c) Only a constant may be pulled out of an integral. Correct: $x \sin x + \cos x + C$ .

(d) The numerator is not the derivative of the denominator. Correct: $\arctan x + C$ .

Solution

Each error can be exposed by the same test: differentiate the proposed answer and see whether the integrand comes back.

(a) The power rule requires

α \neq - 1

, since

α + 1 = 0

makes the denominator zero. The correct antiderivative is the logarithm:

\int \frac{d x}{x} = \ln | x | + C,

since $\frac{d}{d x} \ln | x | = \frac{1}{x}$ .

(b) There is no product rule for integration. Differentiating the proposed answer gives

x^{2} \cdot \frac{x^{4}}{4} + \frac{x^{3}}{3} \cdot x^{3}

, which is not

x^{5}

. Combine the powers first:

\int x^{2} \cdot x^{3} d x = \int x^{5} d x = \frac{x^{6}}{6} + C .

(c) Property (a) of the linearity theorem,

\int k f (x) d x = k \int f (x) d x

, holds only when

k

is a constant. Here

x

is a variable, and indeed

\frac{d}{d x} (x \sin x) = \sin x + x \cos x \neq x \cos x

. The correct answer, obtainable by integration by parts or simply by verifying, is

\int x \cos x d x = x \sin x + \cos x + C,

since

\begin{aligned} \frac{d}{dx}\left(x\sin x+\cos x\right) &= \sin x+x\cos x-\sin x \\ &= x\cos x. \end{aligned}
(d) Differentiating the proposed answer gives

\frac{2 x}{x^{2} + 1}

, not

\frac{1}{x^{2} + 1}

. The correct entry is the arctangent formula:

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

Since integration is an indirect process, an antiderivative can always be checked by differentiating it. Verify each of the following formulas, which will be derived later in the course:

(a) $\int \ln x d x = x \ln x - x + C (x > 0)$

(b) $\int \sec x d x = \ln | \sec x + \tan x | + C$

Answer

Both are verified by differentiating the right-hand side and recovering the integrand.

Solution

(a) Apply the Product Rule to

x \ln x

: \begin{aligned} \frac{d}{dx}\left(x\ln x-x\right) &= \left(1\cdot\ln x+x\cdot\frac{1}{x}\right)-1 \\ &= \ln x+1-1 \\ &= \ln x. \end{aligned}

The derivative of the right-hand side is exactly the integrand, so the formula is correct. ✓

(b) Apply the Chain Rule, using \dfrac{d}{dx}\ln|v|=\dfrac{v'}{v} with

v = \sec x + \tan x

: \begin{aligned} \frac{d}{dx}\ln|\sec x+\tan x| &= \frac{\sec x\tan x+\sec^{2}x}{\sec x+\tan x} \\ &= \frac{\sec x\left(\tan x+\sec x\right)}{\sec x+\tan x} \\ &= \sec x. \end{aligned}

The integrand is recovered, so the formula is correct. ✓ This is a good illustration of the theme of this section: finding such an antiderivative is hard, but checking it is routine.

Evaluate the following, combining several of the techniques above:

(a) $\int \frac{(x - 1) (x^{2} + 3)}{x^{2}} d x$

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

(c) $\int \frac{{(x^{3 / 2} + x^{- 3 / 2})}^{2}}{x} d x$

Answer

(a) $\frac{x^{2}}{2} - x + 3 \ln | x | + \frac{3}{x} + C$

(b) $\ln | x | + 2 \arctan x + C$

(c) $\frac{x^{3}}{3} + 2 \ln | x | - \frac{1}{3 x^{3}} + C$

Solution

(a) Multiply out the numerator, then divide by

x^{2}

\frac{x^{3} - x^{2} + 3 x - 3}{x^{2}} = x - 1 + \frac{3}{x} - \frac{3}{x^{2}} .

Integrating term by term,

\begin{aligned} \int \frac{(x-1)\left(x^{2}+3\right)}{x^{2}}\,dx &= \frac{x^{2}}{2}-x+3\ln|x| \\ &\quad -3\cdot\frac{x^{-1}}{-1}+C \\ &= \frac{x^{2}}{2}-x+3\ln|x|+\frac{3}{x}+C. \end{aligned}
(b) Expand the numerator and group the terms so that

1 + x^{2}

appears: \begin{aligned} \frac{(1+x)^{2}}{x\left(1+x^{2}\right)} &= \frac{1+2x+x^{2}}{x\left(1+x^{2}\right)} \\ &= \frac{\left(1+x^{2}\right)+2x}{x\left(1+x^{2}\right)} \\ &= \frac{1}{x}+\frac{2}{1+x^{2}}. \end{aligned}

Therefore

\begin{aligned} \int \frac{(1+x)^{2}}{x\left(1+x^{2}\right)}\,dx \\ = \ln|x|+2\arctan x+C. \end{aligned}
(c) Expand the square, remembering the middle term

2 \cdot x^{3 / 2} \cdot x^{- 3 / 2} = 2

{(x^{3 / 2} + x^{- 3 / 2})}^{2} = x^{3} + 2 + x^{- 3},

so after dividing by $x$ the integrand is $x^{2} + \frac{2}{x} + x^{- 4}$ , and

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

Integration Is an Indirect Process

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