Computing integrals (3): Integration by parts

 Sometimes, we may simplify integration by using the product rule of differentiation. This technique is called integration by parts.

Theorem (Integration by parts)

Let $f(x)$ and $g(x)$ be differentiable functions on an open interval $I$. Then,

1.  $\int f(x)g^{\prime }(x)dx=f(x)g(x)-\int f^{\prime }(x)g(x)dx$;
2. For any $a,b\in I$, $${\int }_a^{b}f(x)g^{\prime }(x)dx={[f(x)g(x)]}_a^b-{\int }_a^b{f}^{\prime }(x)g(x)dx.$$

Proof. By the product rule,

$$[f(x)g(x){]}^{\prime }=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)$$

so

$$f(x)g^{\prime }(x)=[f(x)g(x){]}^{\prime }-f^{\prime }(x)g(x).$$

By integrating both sides, we have the desired results. ■

Example. Let us find $\int x\cosh xdx$.

$$\begin{array}{rcl}\int x\cosh xdx& =& \int x(\sinh x{)}^{\prime }dx\\ & =& x\sinh x-\int 1\cdot \sinh xdx\\ & =& x\sinh x-\cosh x+C.\end{array}$$

Example (eg:recur). Let us study how we can compute

$$I_n=\int \frac{dx}{(x^2+1{)}^n}$$

for $n\in \mathbb{N}$. Note

$$I_n=\int \frac{x^2+1}{(x^2+1{)}^{n+1}}dx=\int \frac{x^2}{(x^2+1{)}^{n+1}}dx+I_{n+1}.$$

Since

$${(-\frac{1}{2n(x^2+1{)}^n})}^{\prime }=\frac{x}{(x^2+1{)}^{n+1}},$$

we have

$$\begin{array}{rcl}\int \frac{x^2}{(x^2+1{)}^{n+1}}dx& =& \int x\cdot \frac{x}{(x^2+1{)}^{n+1}}dx\\ & =& -\frac{x}{2n(x^2+1{)}^n}+\frac{1}{2n}\int \frac{dx}{(x^2+1{)}^n}\\ & =& \frac{1}{2n}{I}_n-\frac{x}{2n(x^2+1{)}^n}.\end{array}$$

Therefore, we have the following recurrence equation

$$\begin{array}{rcl}{I}_{n+1}& =& I_n-\int \frac{x^2}{(x^2+1{)}^{n+1}}dx\\ \text{(eq:Irec)}& & =& (1-\frac{1}{2n})I_n+\frac{1}{2n}\cdot \frac{x}{(x^2+1{)}^n}.\end{array}$$

Starting from

$$I_1=\int \frac{dx}{x^2+1}=\arctan x+C,$$

we can compute $I_2,I_3,\cdots$ recursively by using Eq. (eq:Irec).