7.4: Integration by Parts

One reason it is harder to integrate than differentiate is that for derivatives there is both a Sum Rule and a Product Rule,

$$d(u+v)=d u+d v, \quad d(u v)=u d v+v d u \nonumber$$

while for integrals there is only a Sum Rule,

$$\int d u+d v=\int d u+\int d v \nonumber$$

The Sum Rule for integrals is obtained in a simple way by reversing the sum rule for derivatives.

There is a way to turn the Product Rule for derivatives into a rule for integrals. It no longer looks like a product rule, and is called integration by parts. Integration by parts is a basic method which is needed for many integrals involving trigonometric functions (and later exponential functions).

Integration by Parts

Suppose, for $x$ in an open interval $I$, that $u$ and $v$ depend on $x$ and that du and $d v$ exist. Then

$$\int u d v=u v-\int v d u \nonumber$$

PROOF We use the Product Rule

$$u d v+v d u=d(u v), \quad u d v=d(u v)-v d u \nonumber$$

Integrating both sides with $x$ as the independent variable,

$$\int u d v=\int(d(u v)-v d u)=\int d(u v)-\int v d u=u v-\int v d u \nonumber$$

No constant of integration is needed because there are indefinite integrals on both sides of the equation.

Integration by parts is useful whenever $\int v d u$ is easier to evaluate than a given integral $\int u d v$.

Example 1

Evaluate $\int x \sin x d x$. Our plan is to break $x \sin x d x$ into a product of the form $u d v$, evaluate the integrals $\int d v$ and $\int v d u$, and then use integration by parts to get $\int u d c$. There are several choices we might make for $u$ and $d i$, and not all of them lead to a solution of the problem. Some guesswork is required.

Solution

First $\operatorname{tr} y: u=\sin x, d v=x d x . \int d v=\int x d x=\frac{1}{2} x^{2}+C$. Take $v=\frac{1}{2} x^{2}$. Next we find $d u$ and try to evaluate $\int v d u$.

$$d u=\cos x d x, \quad \int v d u=\int \frac{1}{2} x^{2} \cos x d x \nonumber$$

This integral looks harder than the one we started with, so we shall start over with another choice of $u$ and $d v$.

Second try: $u=x, d u=\sin x d x$.

$$\int d v=\int \sin x d x=-\cos x+C \nonumber$$

We take $v=-\cos x$. This time we find $d u$ and easily evaluate $\int v d u$.

$$d u=d x, \quad \int u d u=\int-\cos x d x=-\sin x+C_{1} \nonumber$$

Finally we use the rule

$$\begin{aligned} \int u d v & =u v-\int v d u \\ \int x \sin x d x & =x(-\cos x)-\left(-\sin x+C_{1}\right) \\ \int x \sin x d x & =-x \cos x+\sin x+C \end{aligned} \nonumber$$

Example 2

Evaluate $\int \arcsin x d x$. A choice of $u$ and $d v$ which works is

Solution

$$u=\arcsin x, \quad d v=d x \nonumber$$

We may take $v=x$. Then

$$\begin{aligned} d u & =\frac{d x}{\sqrt{1-x^{2}}} \\ \int v d u & =\int-\frac{x d x}{\sqrt{1-x^{2}}}=-\sqrt{1-x^{2}}+C_{1} \end{aligned} \nonumber$$

Finally, $\quad \int \arcsin x d x=x \arcsin x-\left(-\sqrt{1-x^{2}}+C_{1}\right)$,

$$\int \arcsin x d x=x \arcsin x+\sqrt{1-x^{2}}+C \nonumber$$

This integral and the similar formula for $\int \arccos x d x$ are included in our table at the end of the book. We shall see how to integrate the other inverse trigonometric functions in the next chapter.

Example 3

Evaluate $\int x^{2} \sin x d x$. This requires two integrations by parts.

Solution

Step 1

$$\begin{aligned} u & =x^{2}, & d v & =\sin x d x \\ d u & =2 x d x, & \int d v & =\int \sin x d x=-\cos x+C \end{aligned} \nonumber$$

We take $v=-\cos x$.

$$\int x^{2} \sin x d x=u v-\int v d u=-x^{2} \cos x+\int 2 x \cos x d x \nonumber$$

Step 2 Evaluate $\int 2 x \cos x d x$.

$$\begin{array}{rlrl} u_{1} & =2 x, & d v_{1} & =\cos x d x \\ d u_{1} & =2 d x, \quad \int d v_{1} & =\int \cos x d x=\sin x+C \end{array} \nonumber$$

We take $v_{1}=\sin x$.

$$\begin{aligned} \int 2 x \cos x d x & =u_{1} v_{1}-\int v_{1} d u_{1} \\ & =2 x \sin x-\int 2 \sin x d x \\ & =2 x \sin x+2 \cos x+C . \end{aligned} \nonumber$$

Combining the two steps,

$$\int x^{2} \sin x d x=-x^{2} \cos x+2 x \sin x+2 \cos x+C \nonumber$$

Sometimes integration by parts will yield an equation in which the given integral occurs on both sides. One can often solve for the answer.

Example 4

Evaluate $\int \sin ^{2} \theta d \theta$. Let

Solution

Then

$$\begin{aligned} u & =\sin \theta, & d v & =\sin \theta d \theta \\ d u & =\cos \theta d \theta, & v & =-\cos \theta \end{aligned} \nonumber$$

$$\begin{aligned} \int \sin ^{2} \theta d \theta & =-\sin \theta \cos \theta-\int-\cos ^{2} \theta d \theta \\ & =-\sin \theta \cos \theta+\int \cos ^{2} \theta d \theta \\ & =-\sin \theta \cos \theta+\int\left(1-\sin ^{2} \theta\right) d \theta \\ & =-\sin \theta \cos \theta+\theta-\int \sin ^{2} \theta d \theta \end{aligned} \nonumber$$

We solve this equation for $\int \sin ^{2} \theta d \theta$,

$$\int \sin ^{2} \theta d \theta=-\frac{1}{2} \sin \theta \cos \theta+\frac{1}{2} \theta+C \nonumber$$

Here is another way to evaluate $\int \sin ^{2} \theta d \theta$. Instead of using integration by parts, we can use the half-angle formula

$$\sin ^{2} \theta=\frac{1-\cos (20)}{2} \nonumber$$

This is derived from the addition formula,

$$\begin{aligned} \cos (\theta+\phi) & =\cos \theta \cos \phi-\sin \theta \sin \phi \\ \cos (2 \theta) & =\cos ^{2} \theta-\sin ^{2} \theta=1-2 \sin ^{2} \theta \\ \sin ^{2} \theta & =\frac{1-\cos (2 \theta)}{2} \end{aligned} \nonumber$$

Then

$$\begin{aligned} \int \sin ^{2} \theta d \theta & =\int \frac{1-\cos 2 \theta}{2} d \theta=\frac{1}{2} \int d \theta-\frac{1}{2} \int \cos 2 \theta d \theta \\ & =\frac{1}{2} \int d \theta-\frac{1}{4} \int \cos 2 \theta d(2 \theta)=\frac{1}{2} \theta-\frac{1}{4} \sin 2 \theta+C \end{aligned} \nonumber$$

This answer agrees with Example 4 because

so

$$\begin{aligned} & \sin 2 \theta=\sin (\theta+\theta)=2 \sin \theta \cos \theta \\ & \frac{1}{2} \theta-\frac{1}{4} \sin 2 \theta=\frac{1}{2} \theta-\frac{1}{2} \sin \theta \cos \theta \end{aligned} \nonumber$$

Integration by parts requires a great deal of guesswork. Given a problem $\int h(x) d x$ we try to find a way to split $h(x) d x$ into a product $f(x) g^{\prime}(x) d x$ where we can evaluate both of the integrals $\int g^{\prime}(x) d x$ and $\int g(x) f^{\prime}(x) d x$.

Definite integrals take the following form when integration by parts is applied.

DEFINITE INTEGRATION BY PARTS

If $u=f(x)$ and $x=g(x)$ have continuous derivatives on an open interval $I$, then for $a, b$ in $I$,

$$\left.\int_{a}^{b} f(x) g^{\prime}(x) d x=f(x) g(x)\right]_{a}^{b}-\int_{a}^{b} g(x) f^{\prime}(x) d x \nonumber$$

PROOF The Product Rule gives

$$f(x) g^{\prime}(x) d x+g(x) f^{\prime}(x) d x=d(f(x) g(x)) \nonumber$$

Then by the Fundamental Theorem of Calculus,

$$\left.\int_{a}^{b}\left(f(x) g^{\prime}(x)+g(x) f^{\prime}(x)\right) d x=f(x) g(x)\right]_{a}^{b} \nonumber$$

and the desired result follows by the Sum Rule.

If we plot $u=f(x)$ on one axis and $v=g(x)$ on the other, we get a picture of definite integration by parts (Figure 7.4.1). The picture is easier to interpret if we change variables in the definite integrals and write the formula for integration by parts in the form

$$\int_{g(a)}^{g(b)} u d v+\int_{f(a)}^{f(b)} v d u=f(b) g(b)-f(a) g(a) \nonumber$$

Figure 7.4.1 Definite Integration by Parts

Example 5

Evaluate $\int_{0}^{\pi} x \sin x d x$ (Figure 7.4.2). Take $u=x, d v=\sin x d x$ as in Example 1. Then $v=-\cos x$ and

$$\begin{aligned} \int_{0}^{\pi} x \sin x d x & =-x \cos x]_{0}^{\pi}-\int_{0}^{\pi}-\cos x d x \\ & \left.=-x \cos x]_{0}^{\pi}+\sin x\right]_{0}^{\pi} \\ & =(-\pi(-1)+0 \cdot 1)+(0-0)=\pi \end{aligned} \nonumber$$

Solution

 

 

Figure 7.4.2

 

PROBLEMS FOR SECTION 7.4

Evaluate the integrals in Problems 1-35.
$1 \int x \cos x d x$
$2 \int \arccos x d x$
$3 \int t^{2} \cos t d t$
$4 \int x \arctan x d x$
$5 \int t \sin (2 t-1) d t$
$6 \int \arcsin (3 t) d t$
$7 \quad \int x^{2} \sin (4 x) d x$
$8 \quad \int x \operatorname{arcsec} x d x$
$9 \int x^{3} \operatorname{arcsec} x d x$
$10 \int x^{3} \sin x d x$
$11 \int \sin \sqrt{x} d x$
$12 \int \sin \theta \tan ^{2} \theta d \theta$
$13 \int \arctan x d x$
$14 \int x \tan x \sec ^{2} x d x$
15
$\int \frac{x^{3}}{\sqrt{x^{2}-1}} d x$
$16 \int \cos ^{2} \theta d \theta$
$17 \int x \sin x \cos x d x$
$18 \int t \sin ^{2} t d t$
19
$\int \sin \theta \sin (2 \theta) d \theta$
20 $\int \cos x \cos (3 x) d x$
$21 \int \sin x \cos (5 x) d x$
$22 \int \cos x \cot ^{4} x d x$
$23 \int t^{3} \sin \left(t^{2}\right) d t$
$24 \int x^{3} \cos \left(2 x^{2}-1\right) d x$
25
$\int \frac{1}{x^{3}} \sin \left(\frac{1}{x}\right) d x$
26
$\int \sin \theta \cos \theta \cos (\sin \theta) d \theta$
$27 \int t^{3} \sqrt{t^{2}+4} d t$
$28 \int \frac{1}{x^{3}} \sqrt{\frac{1}{x}-1} d x$
29
$\int_{0}^{\pi \cdot 2} \theta \cos \theta d \theta$
$30 \quad \int_{0}^{1.2} \arcsin x d x$
$31 \int_{0}^{\pi} \sin ^{2} \theta d \theta$
$32 \int_{0}^{1} \arcsin x d x$
33
$\int_{0}^{1} x \operatorname{arccot} x d x$
$34 \int_{0}^{x} x \operatorname{arccot} x d x$

36 Find the volume of the solid of revolution generated by rotating the region under the curve $y=\sin x, 0 \leq x \leq \pi$, about (a) the $x$-axis, (b) the $y$-axis.

37 Prove that if $f$ is a differentiable function of $x$, then

$$\int f(x) d x=x f(x)-\int x f^{\prime}(x) d x \nonumber$$

38 If $u$ and $v$ are differentiable functions of $x$, show that

$$\int u^{2} d u=u^{2} v-2 \int u d u \nonumber$$

39 Show that if $f^{\prime}$ and $g$ are differentiable for all $x$, then

$$\int g(x) g^{\prime}(x) f^{\prime \prime}(g(x)) d x=f^{\prime}(g(x)) g(x)-f(g(x))+C \nonumber$$