Basic Integration Rules
In this section, we will study several integration techniques for fitting an integrand into one of the basic integration rules. The basic integration rules are reviewed in Table 6.1 on page 252 .
Procedures for Fitting Integrands to Basic Rules
Procedure Example
1. Separating numerator $\frac{1-2 x}{1+x^2}=\frac{1}{1+x^2}-\frac{2 x}{1+x^2}$
2. Adding and subtracting terms in numerator $\frac{1}{1-e^x}=\frac{1-e^x+e^x}{1-e^x}=\frac{1-e^x}{1-e^x}+\frac{e^x}{1-e^x}$
3. Dividing improper fractions $\frac{x^3-3 x}{x^2-1}=x-\frac{2 x}{x^2-1}$
4. Completing the square $\frac{1}{\sqrt{4 x-x^2}}=\frac{1}{\sqrt{4-(x-2)^2}}$
Other integration techniques, such as the simple substitution method, were covered in section 4.8. Using trigonometric identities, trigonometric substitution, Method of Partial Fractions and Integration by Parts will be covered later in this chapter.
1. Separating numerator
Example 1
- Evaluate $\int \frac{1-2 x}{1+x^2} d x$.
Answer/Explanation
Solution
$\int \frac{1-2 x}{1+x^2} d x=\int \frac{1}{1+x^2} d x+\int \frac{-2 x}{1+x^2} d x$ Separate the numerator
$\int \frac{1}{1+x^2} d x=\arctan x$ Basic integration rules
$\int \frac{-2 x}{1+x^2} d x=\int \frac{-d u}{u}=-\ln u$ Let $u=1+x^2$, then $d u=2 x d x$.
Therefore $\int \frac{1-2 x}{1+x^2} d x=\arctan x-\ln \left(1+x^2\right)+C$.
Differentiation Rules and Basic Integration Rules
2. Adding and subtracting terms in numerator
Example 2
- Evaluate $\int \frac{1}{1-e^x} d x$.
Answer/Explanation
Solution
$\int \frac{1}{1-e^x} d x=\int \frac{1-e^x+e^x}{1-e^x} d x$ Add and subtract $e^x$ in the numerator.
$=\int \frac{1-e^x}{1-e^x} d x+\int \frac{e^x}{1-e^x} d x$ Separate the numerator
$=\int d x+\int \frac{e^x}{1-e^x} d x$
$=x-\ln \left(1-e^x\right)+C$ Use the basic integration rules.
3. Dividing improper fractions
Example 3
- Evaluate $\int \frac{x^3-3 x}{x^2-1} d x$.
Answer/Explanation
Solution
$
\int \frac{x^3-3 x}{x^2-1} d x=\int\left(x-\frac{2 x}{x^2-1}\right) d x
$ Divide an improper fraction.
$=\int x d x-\int \frac{2 x}{x^2-1} d x$
$=\frac{1}{2} x^2-\ln \left(x^2-1\right)+C $ Use the basic integration rules.
Example4
- $
\int \frac{1+\sin x}{\cos ^2 x} d x=
$
(A) $\tan x-\sec x \tan x+C$
(B) $\tan x+\sec x+C$
(C) $\tan x+\sec ^2 x+C$
(D) $\ln \left(1+\cos ^2 x\right)+C$
▶️Answer/Explanation
Ans:B
Example5
- $
\int 2 \tan x \ln (\cos x) d x=
$
(A) $\cos x[\ln (\cos x)]+C$
(B) $\sin x[\ln (\cos x)]+C$
(C) $-[\ln (\cos x)]^2+C$
(D) $[\ln (\sin x)]^2+C$
▶️Answer/Explanation
Ans:D
Example6
- $
\int 2 \tan x \ln (\cos x) d x=
$
(A) $\cos x[\ln (\cos x)]+C$
(B) $\sin x[\ln (\cos x)]+C$
(C) $-[\ln (\cos x)]^2+C$
(D) $[\ln (\sin x)]^2+C$
▶️Answer/Explanation
Ans:C
6.2 Trigonometric Integrals
Trigonometric Identities
$
\begin{aligned}
& \sin ^2 x+\cos ^2 x=1 \quad \tan ^2 x+1=\sec ^2 x \quad \cot ^2 x+1=\csc ^2 x \\
& \sin ^2 x=\frac{1-\cos 2 x}{2} \quad \cos ^2 x=\frac{1+\cos 2 x}{2} \\
& \sin 2 x=2 \sin x \cos x
\end{aligned}
$
Guidelines for Evaluating $\int \sin ^m x \cos ^n x d x$.
1. If $m$ is odd, save one sine factor and use $\sin ^2 x=1-\cos ^2 x$ to express the remaining factor in terms of cosine.
Example 1
- Evaluate $\int \sin ^3 x \cos ^2 x d x$
Answer/Explanation
Solution
$\begin{array}{lll}=\int \sin ^2 x \cos ^2 x(\sin x) d x & & \text { One sine factor is saved. } \\ =\int\left(1-\cos ^2 x\right) \cos ^2 x(\sin x) d x & & \sin ^2 x=1-\cos ^2 x \\ =\int\left(\cos ^2 x-\cos ^4 x\right) \sin x d x & & \text { Multiply. } \\ =\int\left(u^2-u^4\right)(-d u) & & u=\cos x, d u=-\sin x d x \\ =-\frac{1}{3} \cos ^3 x+\frac{1}{5} \cos ^5 x+C & \end{array}$
2. If $n$ is odd, save one cosine factor and use $\cos ^2 x=1-\sin ^2 x$ to express the remaining factor in terms of sine.
Example 2
- Evaluate $\int \cos ^5 x d x$
Answer/Explanation
Solution
$\int \cos ^5 x d x$
$=\int \cos ^4 x(\cos x) d x$ One cosine factor is saved.
$=\int\left(1-\sin ^2 x\right)^2 \cos x d x \quad \cos ^2 x=1-\sin ^2 x$
$=\int\left(1-u^2\right)^2 d u \quad u=\sin x, d u=\cos x d x$
$=\int\left(1-2 u^2+u^4\right) d u$ Multiply.
$\begin{aligned} & =u-\frac{2}{3} u^3+\frac{1}{5} u^5+C \\ & =\sin x-\frac{2}{3} \sin ^3 x+\frac{1}{5} \sin ^5 x+C\end{aligned}$
Example3
- If both $m$ and $n$ are even, substitute $\sin ^2 x=\frac{1-\cos 2 x}{2}$ and $\cos ^2 x=\frac{1+\cos 2 x}{2}$ to reduce the integrand to lower powers of $\cos 2 x$.
Evaluate $\int \sin ^2 x \cos ^2 x d x$.
▶️Answer/Explanation
Solution
$
\begin{aligned}
& \int \sin ^2 x \cos ^2 x d x \\
& =\int\left(\frac{1-\cos 2 x}{2}\right)\left(\frac{1+\cos 2 x}{2}\right) d x \quad \sin ^2 x=\frac{1-\cos 2 x}{2}, \cos ^2 x=\frac{1+\cos 2 x}{2} \\
& =\frac{1}{4} \int\left(1-\cos ^2 2 x\right) d x \\
& =\frac{1}{4} \int\left(1-\frac{1+\cos 4 x}{2}\right) d x \quad \cos ^2 2 x=\frac{1+\cos 4 x}{2} \\
& =\frac{1}{8} \int(1-\cos 4 x) d x \\
& =\frac{1}{8} x-\frac{1}{32} \sin 4 x+C \\
&
\end{aligned}
$
Example5
- $
\int \sin ^3 n x d x=
$
(A) $\frac{1}{3 n} \sin ^3 n x-\frac{1}{n} \sin n x+C$
(B) $\frac{1}{3 n} \cos ^3 n x-\frac{1}{n} \cos n x+C$
(C) $\frac{1}{3 n} \sin ^3 n x-\frac{1}{n} \sin n x+C$
(D) $\frac{1}{3 n} \cos ^3 n x-\frac{1}{n} \sin n x+C$
▶️Answer/Explanation
Ans:B
Example6
- $
\int \cos ^3 x \sqrt{\sin x} d x=
$
(A) $\frac{1}{3}(\cos x)^3-\frac{2}{5}(\cos x)^{5 / 2}+C$
(B) $\frac{2}{3}(\cos x)^{3 / 2}-\frac{2}{7}(\sin x)^{7 / 2}+C$
(C) $\frac{2}{3}(\sin x)^{3 / 2}-\frac{2}{7}(\sin x)^{7 / 2}+C$
(D) $\frac{2}{3}(\sin x)^{3 / 2}-\frac{2}{5}(\cos x)^{5 / 2}+C$
▶️Answer/Explanation
Ans:C
6.3 Trigonometric Substitutions
Trigonometric Substitution
1. For integrals involving $\sqrt{a^2-u^2}$, let $u=a \sin \theta$. Then
$
\sqrt{a^2-u^2}=\sqrt{a^2-a^2 \sin ^2 \theta}=\sqrt{a^2\left(1-\sin ^2 \theta\right)}=\sqrt{a^2 \cos ^2 \theta}=a \cos \theta
$
2. For integrals involving $\sqrt{a^2+u^2}$, let $u=a \tan \theta$. Then
$
\sqrt{a^2+u^2}=\sqrt{a^2+a^2 \tan ^2 \theta}=\sqrt{a^2\left(1+\tan ^2 \theta\right)}=\sqrt{a^2 \sec ^2 \theta}=a \sec \theta
$
3. For integrals involving $\sqrt{u^2-a^2}$, let $u=a \sec \theta$. Then
$
\sqrt{u^2-a^2}=\sqrt{a^2 \sec ^2 \theta-a^2+}=\sqrt{a^2\left(\sec ^2 \theta-1\right)}=\sqrt{a^2 \tan ^2 \theta}=a \tan \theta
$
Note: $\operatorname{arcsec} x=\arccos \frac{1}{x}$
$
\operatorname{arccsc} x=\arcsin \frac{1}{x}
$
Example 1
- Evaluate $\int_0^2 \sqrt{4-x^2} d x$.
Answer/Explanation
Solution
$\begin{aligned} & x=2 \sin \theta, d x=2 \cos \theta d \theta \\ & \sqrt{4-4 \sin ^2 \theta}=\sqrt{4\left(1-\sin ^2 \theta\right)}=2 \cos \theta \\ & \cos ^2 \theta=\frac{1+\cos 2 \theta}{2}\end{aligned}$
$\begin{aligned} & =2 \sin ^{-1}\left(\frac{x}{2}\right)+2\left(\frac{x}{2}\right)\left(\frac{\sqrt{4-x^2}}{2}\right)+C \\ & =2 \sin ^{-1}\left(\frac{x}{2}\right)+\frac{x \sqrt{4-x^2}}{2}+C\end{aligned}$
$\begin{aligned} & \int \sqrt{4-x^2} d x \\ & =\int \sqrt{4-4 \sin ^2 \theta} 2 \cos \theta d \theta \\ & =4 \int \cos ^2 \theta d \theta \\ & =4 \int \frac{1+\cos 2 \theta}{2} d \theta \\ & =2 \int(1+\cos 2 \theta) d \theta \\ & =2\left(\theta+\frac{1}{2} \sin 2 \theta\right)+C=2 \theta+\sin 2 \theta+C \\ & =2 \theta+2 \sin \theta \cos \theta+C \\ & =2 \sin ^{-1}\left(\frac{x}{2}\right)+2\left(\frac{x}{2}\right)\left(\frac{\sqrt{4-x^2}}{2}\right)+C \\ & =2 \sin ^{-1}\left(\frac{x}{2}\right)+\frac{x \sqrt{4-x^2}}{2}+C \\ & \end{aligned}$
Therefore
$
\int_0^2 \sqrt{4-x^2} d x=\left[2 \sin ^{-1}\left(\frac{x}{2}\right)+\frac{x \sqrt{4-x^2}}{2}\right]_0^2=2 \sin ^{-1}(1)=2 \cdot \frac{\pi}{2}=\pi
$
Example2
- Evaluate $\int \frac{d x}{\sqrt{9+x^2}}$.
Answer/Explanation
Solution
$x=3 \tan \theta, d x=3 \sec ^2 \theta d \theta$
$\begin{aligned} & \int \frac{d x}{\sqrt{9+x^2}} \\ & =\int \frac{3 \sec ^2 \theta d \theta}{\sqrt{9+9 \tan ^2 \theta}} \\ & =\int \frac{3 \sec ^2 \theta d \theta}{3 \sec \theta} \\ & =\int \sec \theta d \theta \\ & =\ln |\sec \theta+\tan \theta|+C \\ & =\ln \left|\frac{\sqrt{9+x^2}}{3}+\frac{x}{3}\right|+C\end{aligned}$
Example 3
- $\begin{aligned} & x=2 \sec \theta, d x=2 \sec \theta \tan \theta d \theta \\ & \sqrt{4 \sec ^2 \theta-4}=\sqrt{4\left(\sec ^2 \theta-1\right)}=2 \tan \theta\end{aligned}$
▶️Answer/Explanation
$\begin{aligned} & \int \frac{d x}{x^2 \sqrt{x^2-4}} \\ & =\int \frac{2 \sec \theta \tan \theta d \theta}{4 \sec ^2 x \sqrt{4 \sec ^2 \theta-4}} \\ & =\int \frac{2 \sec \theta \tan \theta d \theta}{4 \sec ^2 \theta 2 \tan \theta} \\ & =\frac{1}{4} \int \frac{d \theta}{\sec \theta} \\ & =\frac{1}{4} \int \cos \theta d \theta \\ & =\frac{1}{4} \sin \theta+C \\ & =\frac{1}{4} \frac{\sqrt{x^2-4}}{x}+C\end{aligned}$
Example4
- If the substitution $x=2 \tan \theta$ is made in $\int \frac{x^3}{\sqrt{x^2+4}} d x$, where $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$, the resulting integral is
(A) $4 \int \tan ^2 \theta \sec \theta d \theta$
(B) $4 \int \tan ^2 \theta \sec ^2 \theta d \theta$
(C) $8 \int \tan ^3 \theta d \theta$
(D) $8 \int \tan ^3 \theta \sec \theta d \theta$
▶️Answer/Explanation
Ans:D
Example5
- $
\int_{\sqrt{2}}^2 \frac{1}{x \sqrt{x^2-1}} d x=
$
(A) $\frac{\pi}{18}$
(B) $\frac{\pi}{12}$
(C) $\frac{\pi}{6}$
(D) $\frac{\pi}{4}$
▶️Answer/Explanation
Ans:B
Example6
- $
\int \frac{1}{x^2 \sqrt{25-x^2}} d x=
$
(A) $-\frac{\sqrt{25-x^2}}{5 x^2}+C$
(B) $-\frac{\sqrt{25-x^2}}{25}+C$
(C) $-\frac{\sqrt{25-x^2}}{25 x}+C$
(D) $\frac{\sqrt{25-x^2}}{25 x^2}+C$
▶️Answer/Explanation
Ans:C
Example6
- If the substitution $x=\sec \theta$ is made in $\int \frac{\sqrt{x^2-1}}{x^4} d x$, where $-\frac{\pi}{2}<\theta<\frac{\pi}{2}$, the resulting integral is
(A) $\int \sec ^2 \theta \tan \theta d \theta+C$
(B) $\int \sec \theta \tan ^2 \theta d \theta+C$
(C) $\int \sin \theta \cos ^2 \theta d \theta+C$
(D) $\int \sin ^2 \theta \cos \theta d \theta+C$
▶️Answer/Explanation
Ans:D
6.4 L’Hospital’s Rule
L’Hospital’s Rule
Suppose $f$ and $g$ are differentiable and $g^{\prime}(x) \neq 0$ near $x=c$ (except possibly at $c$ ).
If the limit of $\frac{f(x)}{g(x)}$ as $x$ approaches $c$ produces the indeterminate form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then
$
\lim _{x \rightarrow c} \frac{f(x)}{g(x)}=\lim _{x \rightarrow c} \frac{f^{\prime}(x)}{g^{\prime}(x)}
$
provided the limit on the right side exists.
L’Hospital’s Rule can be applied only to quotients leading to indeterminate forms such as
$\frac{0}{0}, \frac{\infty}{\infty}, 0 \cdot \infty, 1^{\infty}, \infty^0, 0^0$, and $\infty-\infty$.
L’Hospital’s Rule does not apply when either the numerator or denominator has a finite nonzero limit.
Example 1
- Find $\lim _{x \rightarrow 0} \frac{e^x-1}{\sin x}$
Answer/Explanation
Solution
Indeterminate form $\frac{0}{0}$
L’Hospital’s Rule: $\frac{d}{d x}\left(e^x-1\right)=e^x, \frac{d}{d x}(\sin x)=\cos x$. $e^0=1$ and $\cos 0=1$
$e^0=1$ and $\cos 0=1$.
$\begin{aligned} & \lim _{x \rightarrow 0} \frac{e^x-1}{\sin x} \\ & =\lim _{x \rightarrow 0} \frac{e^x}{\cos x} \\ & =1\end{aligned}$
Example 2
- Find $\lim _{x \rightarrow \pi / 2} \frac{\sec x+9}{\tan x}$.
Answer/Explanation
Solution
$\lim _{x \rightarrow \pi / 2} \frac{\sec x+9}{\tan x}$ Indeterminate form $\frac{\infty}{\infty}$.
$=\lim _{x \rightarrow \pi / 2} \frac{\sec x \tan x}{\sec ^2 x}$ L’Hospital’s Rule: $\frac{d}{d x}(\sec x+9)=\sec x \tan x$,
$\frac{d}{d x}(\tan x)=\sec ^2 x$
$=\lim _{x \rightarrow \pi / 2} \frac{\tan x}{\sec x}$ Simplify
$\begin{aligned} & =\lim _{x \rightarrow \pi / 2} \sin x \\ & =1\end{aligned}$
Example 3
- Find $\lim _{x \rightarrow \infty} x \tan \frac{1}{x}$.
▶️Answer/Explanation
Solution
$
\begin{array}{ll}
\lim _{x \rightarrow 0} \frac{e^x-1}{\sin x} & \text { Indeterminate form } \frac{0}{0} \\
=\lim _{x \rightarrow 0} \frac{e^x}{\cos x} & \text { L’Hospital’s Rule: } \frac{d}{d x}\left(e^x-1\right)=e^x, \frac{d}{d x}(\sin x)=\cos x . \\
=1 & e^0=1 \text { and } \cos 0=1 .
\end{array}
$
Example4
- $
\lim _{x \rightarrow 0} \frac{e^x-1-x}{x^2}=
$
(A) 0
(B) $\frac{1}{2}$
(C) 1
(D) $\infty$
▶️Answer/Explanation
Ans:B
Example5
- $
\lim _{x \rightarrow 0} \frac{\sin ^{-1} x}{x}=
$
(A) $-\infty$
(B) 0
(C) $\frac{\pi}{2}$
(D) 1
▶️Answer/Explanation
Ans:D
Example5
$
\lim _{\theta \rightarrow \pi} \frac{\sin \theta}{\theta-\pi}=
$
(A) -1
(B) $-\frac{1}{2}$
(C) 0
(D) $\frac{1}{2}$
▶️Answer/Explanation
Ans:A
6.5 Integration by Partial Fractions
A method for rewriting a rational function into the sum of simpler rational functions is called the method of partial fractions. For instance, the rational function $\frac{5 x+1}{x^2+x-2}$ can be written
as $\frac{5 x+1}{x^2+x-2}=\frac{2}{x-1}+\frac{3}{x+2}$.
We call the fractions $\frac{2}{x-1}$ and $\frac{3}{x+2}$ partial fractions because their denominators are only part of the original denominator. To integrate the rational function $\frac{5 x+1}{x^2+x-2}$, we simply sum the integrals of the partial fractions.
$
\int \frac{5 x+1}{x^2+x-2} d x=\int \frac{2}{x-1} d x+\int \frac{3}{x+2} d x=2 \ln |x-1|+3 \ln |x+2|+C
$
Example 1
- Evaluate $\int \frac{x^3}{x^2-1} d x$.
Answer/Explanation
Solution
If the degree of the numerator is greater than or equal to the degree of the denominator, divide numerator by the denominator to get a polynomial plus a proper fraction.
Then we write the improper fraction as a polynomial plus a proper fraction.
$
\frac{x^3}{x^2-1}=x+\frac{x}{x^2-1}
$
Write $\frac{x}{x^2-1}=\frac{x}{(x+1)(x-1)}=\frac{A}{x+1}+\frac{B}{x-1}$.
Multiplying both sides by $(x+1)(x-1)$, we have $x=A(x-1)+B(x+1)$
Let $x=1$. Then $1=A(1-1)+B(1+1) \Rightarrow B=\frac{1}{2}$.
Let $x=-1$. Then $-1=A(-1-1)+B(-1+1) \Rightarrow A=\frac{1}{2}$.
Thus, $\int \frac{x^3}{x^2-1} d x=\int x d x+\frac{1}{2} \int \frac{1}{x+1} d x+\frac{1}{2} \int \frac{1}{x-1} d x$ $=\frac{1}{2} x^2+\frac{1}{2} \ln |x+1|+\frac{1}{2} \ln |x-1|+C$.
Example 2
- Evaluate $\int \frac{x+10}{(x-4)(x+3)} d x$
Answer/Explanation
Solution
If the denominator is a product of distinct linear factors, then the partial fraction decomposition has the form
$
\frac{x+10}{(x-4)(x+3)}=\frac{A}{x-4}+\frac{B}{x+3}
$
To find the values of $A$ and $B$, multiply both sides of the equation by the least common denominator and get $x+10=A(x+3)+B(x-4)$
To solve for $A$, choose $x=4$, to eliminate the term $B(x-4)$.
$
4+10=A(4+3)+B(4-4) \Rightarrow 14=7 A \Rightarrow A=2
$
To solve for $B$, choose $x=-3$, to eliminate the term $A(x+3)$.
$
\begin{aligned}
& -3+10=A(-3+3)+B(-3-4) \Rightarrow 7=-7 B \Rightarrow B=-1 \\
& \int \frac{x+10}{(x-4)(x+3)} d x=\int \frac{2}{(x-4)} d x+\int \frac{-1}{(x+3)} d x \\
& =2 \ln |x-4|-\ln |x+3|+C \\
&
\end{aligned}
$
Example3
- $
\int \frac{d x}{x^2+x-6}=
$
(A) $\frac{1}{5} \ln \left|\frac{x-1}{x+6}\right|+C$
(B) $\frac{1}{5} \ln \left|\frac{x+3}{x-2}\right|+C$
(C) $\frac{1}{5} \ln \left|\frac{x-2}{x+3}\right|+C$
(D) $\frac{1}{5} \ln |(x-2)(x+3)|+C$
▶️Answer/Explanation
Ans:C
Example4
- $
\int_4^7 \frac{5}{(x-2)(2 x+1)} d x=
$
(A) $\ln \frac{9}{10}$
(B) $\ln \frac{10}{9}$
(C) $\ln \frac{3}{2}$
(D) $\ln \frac{9}{4}$
▶️Answer/Explanation
Ans:C
Example5
- $
\int \frac{x}{x^2+5 x+6} d x=
$
(A) $-2 \ln |x+2|+3 \ln |(x+3)|+C$
(B) $2 \ln |x+2|+3 \ln |(x+3)|+C$
(C) $2 \ln |(x+3)|-3 \ln |x+2|+C$
(D) $-2 \ln |(x+3)|-3 \ln |x+2|+C$
▶️Answer/Explanation
Ans:A
Example6
- $
\int \frac{2 e^{2 x}}{\left(e^x-1\right)\left(e^x+1\right)} d x=
$
(A) $\ln \left|e^x\left(e^{2 x}-1\right)\right|+C$
(B) $\ln \left|2 e^x\left(e^{2 x}-1\right)\right|+C$
(C) $\ln \left|\frac{1}{e^{2 x}-1}\right|+C$
(D) $\ln \left|\left(e^x-1\right)\left(e^x+1\right)\right|+C$
▶️Answer/Explanation
Ans:A
6.6 Integration by Parts $B C$
Integration by Parts Formula
If $u$ and $v$ are functions of $x$ and have continuous derivatives, then
$
\int u d v=u v-\int v d u
$
Guidelines for Integration by Parts
1. For integrals of the form
$\int x^n e^{a x} d x, \int x^n \sin a x d x$, or $\int x^n \cos a x d x$
let $u=x^n$ and let $d v=e^{a x} d x, \sin a x d x$, or $\cos a x d x$.
2. For integrals of the form
$\int x^n \ln x d x, \int x^n \arcsin a x d x, \int x^n \arccos a x d x$, or $\int x^n \arctan a x d x$ let $u=\ln x, \arcsin a x, \arccos a x$, or $\arctan a x$ and let $d v=x^n d x$.
3. For integrals of the form
$\int e^{a x} \sin b x d x$ or $\int e^{a x} \cos b x d x$
let $u=\sin b x$ or $\cos b x$ and let $d v=e^{a x} d x$.
Example1
- Find $\int x \sin x d x$
Answer/Explanation
Solution
Let $u=x$ and $d v=\sin x d x$.
Then $d u=d x$ and $v=\int \sin x d x=-\cos x$.
$
\begin{aligned}
& \int \stackrel{u}{x} \overbrace{\sin x d x}^{d v}=\stackrel{u}{x} \overbrace{(-\cos x)}^v-\int \overbrace{(-\cos x)}^v \overbrace{d x}^{d u} \\
& =-x \cos x+\int \cos x d x \\
& =-x \cos x+\sin x+C \\
&
\end{aligned}
$
Example 2
- Evaluate $\int \arctan x d x$
Answer/Explanation
Solution
Let $u=\arctan x$ and $d v=d x$.
Then $d u=\frac{1}{1+x^2} d x$ and $v=\int d x=x$
$
\begin{aligned}
& \int \overbrace{\arctan x}^u \overbrace{d x}^{d y}=\overbrace{\arctan x}^u \cdot \overbrace{x}^v-\int \frac{v}{x \cdot} \overbrace{\frac{1}{1+x^2} d x}^{d u} \\
& =x \arctan x-\int \frac{x}{1+x^2} d x \\
& =x \arctan x-\frac{1}{2} \ln \left(1+x^2\right)+C
\end{aligned}
$
Tabular Method
In problems involving repeated applications of integration by parts, a tabular method can help to organized the work. This method works well for integrals of the form $\int x^n e^{a x} d x, \int x^n \sin a x d x$, and $\int x^n \cos a x d x$.
Example 3
- Evaluate $\int x^2 e^x d x$
Answer/Explanation
Solution
Let $u=x^2$ and $d v=e^x d x$.
Hence, $\int x^2 e^x d x=x^2 e^x-2 x e^x+2 e^x+C$
Example 4
- Evaluate $\int e^{2 x} \cos x d x$
Answer/Explanation
Solution
Let $u=\cos x$ and $d v=e^{2 x} d x$.
We stop differentiating and integrating as soon as we reach a row that is the same as the first row except for multiplicative constants. The table is interpreted as follow.
$
\int e^{2 x} \cos x d x=\frac{1}{2} e^{2 x} \cos x-\frac{1}{4} e^{2 x}(-\sin x)+\int \overbrace{\frac{1}{4} e^{2 x}(-\cos x)}^{\text {Product of the last row }} d x
$
By adding $\int \frac{1}{4} e^{2 x}(\cos x) d x$ on each side we get,
$
\frac{5}{4} \int e^{2 x} \cos x d x=\frac{1}{2} e^{2 x} \cos x+\frac{1}{4} e^{2 x}(\sin x) .
$
Therefore, $\int e^{2 x} \cos x d x=\frac{2}{5} e^{2 x} \cos x+\frac{1}{5} e^{2 x} \sin x+C$.
Example4
- $
\int x \sin (2 x) d x=
$
(A) $-x \cos (2 x)+\frac{1}{2} \sin (2 x)+C$
(B) $\frac{x}{2} \cos (2 x)-\frac{1}{4} \sin (2 x)+C$
(C) $-\frac{x}{2} \cos (2 x)+\frac{1}{4} \sin (2 x)+C$
(D) $\frac{x}{2} \cos (2 x)+\frac{1}{4} \sin (2 x)+C$
▶️Answer/Explanation
Ans:C
Example4
- $
\int_0^2 x e^x d x=
$
(A) $e^2-1$
(B) $e^2+1$
(C) $e-1$
(D) $e+1$
▶️Answer/Explanation
Ans:B
Example6
- If $\int x^2 \cos (3 x) d x=f(x)-\frac{2}{3} \int x \sin (3 x) d x$, then $f(x)=$
(A) $\frac{2}{3} x \sin (3 x)$
(B) $\frac{1}{3} x^2 \sin (3 x)$
(C) $\frac{2}{3} x \cos (3 x)$
(D) $\frac{1}{3} x \sin (3 x)-\frac{2}{3} \cos (3 x)$
▶️Answer/Explanation
Ans:B
6.7 Improper Integrals $B C$
Improper Integrals with Infinite Integration Limits
1. If $f(x)$ is continuous on $[a, \infty)$, then
$
\int_a^{\infty} f(x) d x=\lim _{b \rightarrow \infty} \int_a^b f(x) d x
$
2. If $f(x)$ is continuous on $(-\infty, b]$, then
$
\int_{-\infty}^b f(x) d x=\lim _{a \rightarrow-\infty} \int_a^b f(x) d x
$
3. If $f(x)$ is continuous on $(-\infty, \infty)$, then
$
\int_{-\infty}^{\infty} f(x) d x=\int_{-\infty}^c f(x) d x+\int_c^{\infty} f(x) d x
$
where $c$ is any real number.
Improper Integrals with Infinite Discontinuities
1. If $f(x)$ is continuous on $[a, b)$ and has an infinite discontinuity at $b$, then
$
\int_a^b f(x) d x=\lim _{c \rightarrow b^{-}} \int_a^c f(x) d x
$
2. If $f(x)$ is continuous on $(a, b]$ and has an infinite discontinuity at $a$, then
$
\int_a^b f(x) d x=\lim _{c \rightarrow a^{+}} \int_c^b f(x) d x
$
3. If $f(x)$ is continuous on $[a, b]$, except for some number $c$ in $(a, b)$ at which $f$ has an infinite discontinuity, then
$
\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x
$
where $c$ is any real number.
In each case, if the limit is finite we say that the improper integral converges and that the limit is the value of the improper integral. If the limit fails to exists, the improper integral diverges.
Example 1
- Evaluate $\int_0^{\infty} x e^{-x^2} d x$
Answer/Explanation
Solution
$
\begin{aligned}
& \int_0^{\infty} x e^{-x^2} d x \\
& =\lim _{b \rightarrow \infty} \int_0^b x e^{-x^2} d x=\lim _{b \rightarrow \infty}\left[-\frac{1}{2} e^{-x^2}\right]_0^b \\
& =-\frac{1}{2} \lim _{b \rightarrow \infty}\left[e^{-b^2}-e^0\right]=-\frac{1}{2}(0-1)=\frac{1}{2}
\end{aligned}
$
Example2
- Evaluate $\int_{-\infty}^{\infty} \frac{d x}{1+x^2}$
Answer/Explanation
Solution
$\begin{aligned} & \int_{-\infty}^{\infty} \frac{d x}{1+x^2}=\int_{-\infty}^0 \frac{d x}{1+x^2}+\int_0^{\infty} \frac{d x}{1+x^2} \\ & =\lim _{a \rightarrow-\infty} \int_a^0 \frac{d x}{1+x^2}+\lim _{b \rightarrow \infty} \int_0^b \frac{d x}{1+x^2} \\ & =\lim _{a \rightarrow-\infty}\left[\tan ^{-1} x\right]_a^0+\lim _{b \rightarrow \infty}\left[\tan ^{-1} x\right]_0^b \\ & =\lim _{a \rightarrow-\infty}\left(\tan ^{-1} 0-\tan ^{-1} a\right)+\lim _{b \rightarrow \infty}\left(\tan ^{-1} b-\tan ^{-1} 0\right) \\ & =\left(0-\left(-\frac{\pi}{2}\right)\right)+\left(\frac{\pi}{2}-0\right)=\pi\end{aligned}$
Example3
- Evaluate $\int_1^5 \frac{d x}{\sqrt{x-1}}$
Answer/Explanation
Solution
$\begin{aligned} & \int_1^5 \frac{d x}{\sqrt{x-1}} \\ & =\lim _{b \rightarrow 1^{+}} \int_b^5 \frac{d x}{\sqrt{x-1}}=\lim _{b \rightarrow 1^{+}}[2 \sqrt{x-1}]_b^5 \\ & =\lim _{b \rightarrow 1^{+}}[2 \sqrt{4}-2 \sqrt{b-1}] \\ & =4\end{aligned}$
Example 5
Find $\int_0^1 \frac{d x}{1-x}$.
Answer/Explanation
Solution
$\begin{aligned} & \int_0^1 \frac{d x}{1-x} \\ & =\lim _{b \rightarrow 1^{-}} \int_0^b \frac{d x}{1-x}=\lim _{b \rightarrow 1^{-}}[-\ln |1-x|]_0^b \\ & =\lim _{b \rightarrow 1^{-}}[-\ln |1-b|+\ln 1] \\ & =\infty\end{aligned}$
Example6
$
\int_2^{\infty} \frac{1}{\sqrt{x-1}} d x=
$
(A) $-\infty$
(B) -2
(C) 1
(D) $\infty$
▶️Answer/Explanation
Ans:D
Example7
- $
\int_0^{\infty} \frac{1}{(x+3)(x+4)} d x=
$
(A) $-\ln \frac{4}{3}$
(B) $-\ln \frac{3}{4}$
(C) 0
(D) $\ln 4$
▶️Answer/Explanation
Ans:B
Example8
- $
\int_0^{\infty} x^2 e^{-x^3}=
$
(A) $\frac{1}{3}$
(B) $\frac{1}{2}$
(C) 1
(D) divergent
▶️Answer/Explanation
Ans:C