Integration by Parts

Tags

Calculus

Cegep/2

Word count

362 words

Reading time

3 minutes

Method of reversing an application of product rule to find the antiderivative
Abbr. IBP

To find the antiderivative of $\int ab\mathrm{d}x$,

1. Set $u$ and $dv$ such that finding $du$ and $v$ is the easiest.
2. Find $du$ by differentiation and $v$ by integration.
3. Substitute $\to uv-\int vdu$.

+ Priority of $u$

1. Logarithmic
2. Inverse trigonometric
3. Algebraic
4. Trigonometric
5. Exponential

Proof

$$\begin{array}{rl}f(x)g(x)& =\int (f^{\prime }(x)g(x)+f(x)g^{\prime }(x))dx\\ \int f(x)g^{\prime }(x)dx& =f(x)g(x)-\int f^{\prime }(x)g(x)dx\end{array}$$

Let $u=f(x)$ and $v=g(x)$, then $du=f^{\prime }(x)dx$ and $dv=g^{\prime }(x)dx$.
Then, by substitution,

$$\int udv=uv-\int vdu$$

Examples

$\int \frac{lnx}{x^2}dx$

Let $u=ln(x)$ and $dv=\frac{1}{x^2}dx$,
then $du=\frac{1}{x}dx$ and $v=-\frac{1}{x}$.

$$\begin{array}{rl}\int \frac{\ln x}{x^2}dx& =-\frac{1}{x}\ln x-\int (-\frac{1}{x})\left(\frac{1}{x}\right)dx\\ & =-\frac{1}{x}\ln x+\int \frac{1}{x^2}dx\\ & =-\frac{1}{x}\ln x-\frac{1}{x}+c\end{array}$$

$\int \arcsin xdx$

Let $u=\arcsin x$ and $dv=dx$,
then $du=\frac{dx}{\sqrt{1-x^2}}$ and $v=x$.
By IBP,

$$\begin{array}{rl}\int \arcsin xdx& =x\arcsin x-\int \frac{x}{\sqrt{1-x^2}}dx\\ & =x\arcsin x+\frac{1}{2}\int \frac{du}{\sqrt{u}}dx\\ & =x\arcsin x+u^{\frac{1}{2}}+c\\ & =x\arcsin x+\sqrt{1-x^2}+c\end{array}$$

$\int {\sec }^{3}xdx$

Let $u=\sec x$ and $dv={\sec }^{2}xdx$,
then $du=\sec x\tan xdx$ and $v=\tan x$.
By IBP,

$$\begin{array}{rl}\int {\sec }^{3}xdx& =\sec x\tan x-\int \sec x{\tan }^{2}xdx\\ & =\sec x\tan x-\int \sec x({\sec }^{2}x-1)dx\\ & =\sec x\tan x-\int {\sec }^{3}xdx+\int \sec xdx\\ 2\int {\sec }^{3}xdx& =\sec x\tan x+\ln |\sec x+\tan x|+c\\ \int {\sec }^{3}xdx& =\frac{1}{2}(\sec x\tan x+\ln |\sec x+\tan x|)+c\end{array}$$

Contributors

Ajitani Hifumi

Changelog

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