# Integration by Parts

Integration by parts is a special technique of integration of two functions when they are multiplied.

Mathematically Integration by parts is given as-

$\int f(x).g(x)dx = f(x)\int g(x)dx-\int f'(x).( \int g(x)dx )dx$

ILATE-

Identify the function that comes first on the following list and select it as f(x).

ILATE stands for-

I : Inverse trigonometric functions : arctan x, arcsec x, arcsin x etc.

L : Logarithmic functions : ln x  , log5(x), etc.

A : Algebraic functions.

T: Trigonometric functions :sinx, cosx , tan x etc .

E: Exponential functions.

Lets Work Out

Examples- Evaluate $\int x.e^{x}dx$

Solution- From ILATE theorem, f(x) = x, and g(x) = $e^{2}$

Thus using the formula for integration by parts, we have

$\int f(x).g(x)dx = f(x)\int g(x)dx-\int f'(x).( \int g(x)dx )dx$

$\int x.e^{x}dx$ = $x.\int e^{x}dx – \int 1. (\int e^{x}dx)dx$

= $x.e^{x} – e^{x} + c$

Example- Evaluate $\int \sqrt{x^{2}- a^{2}}$

Solution- Choosing first function to be $\sqrt{x^{2}- a^{2}}$ and second function to be 1.

$\int \sqrt{x^{2}- a^{2}}$ = $\sqrt{x^{2}- a^{2}}\int 1.dx – \int \frac{1}{2}.\frac{2x}{\sqrt{x^{2}- a^{2}}}.(\int 1.dx).dx$

I = $x.\sqrt{x^{2}- a^{2}} – \int \frac{x^{2}}{\sqrt{x^{2}- a^{2}}}.dx$

Adding and subtracting a2 in the latter part of the integral we have

I = $x.\sqrt{x^{2}- a^{2}} – \int \frac{x^{2}-a^{2}+a^{2}}{\sqrt{x^{2}- a^{2}}}.dx$

I = $x.\sqrt{x^{2}- a^{2}} – \int \frac{x^{2}-a^{2}}{\sqrt{x^{2}- a^{2}}}.dx – \int \frac{a^{2}}{\sqrt{x^{2}- a^{2}}}.dx$

I = $x.\sqrt{x^{2}- a^{2}}$ – I – $a^{2} \int \frac{1}{\sqrt{x^{2}- a^{2}}}.dx$

2I = $x.\sqrt{x^{2}- a^{2}} – a^{2} \log \left | x + \sqrt{x^{2}- a^{2}} \right | + C$

I = $= \frac{x.\sqrt{x^{2}- a^{2}}}{2} – \frac{a^{2}}{2} \log \left | x + \sqrt{x^{2}- a^{2}} \right | + C_{1}$

Example- Evaluate $\int_{0}^{1}\arctan x .dx$

Solution- Let          u = $\arctan x$                    dv = dx

$du = \frac{1}{1+x^{2}}.dx$             v = x

Integration by parts-

$\int_{0}^{1}\arctan x .dx$ = $= \left ( x\arctan x \right )_{0}^{1} – \int_{0}^{1}\frac{x}{1 + x^{2}}dx$

= $\left ( \frac{\pi}{4} – 0 \right ) – \left ( \frac{1}{2} \ln (1+ x^{2}) \right )_{0}^{1}$

= $\left ( \frac{\pi}{4} \right ) – \frac{1}{2} \ln 2$

= $\left ( \frac{\pi}{4} \right ) – \ln \sqrt{2}$