# How do you find the integral of f(x)=( x^2 + 5 ) * cos2x using integration by parts?

Jul 31, 2017

$I = \frac{1}{2} \left({x}^{2} + 5\right) \sin 2 x + \frac{x}{2} \cdot \cos 2 x - \frac{1}{4} \cdot \sin 2 x + C , \mathmr{and} ,$

$I = \frac{1}{4} \left(2 {x}^{2} + 9\right) \sin 2 x + \frac{x}{2} \cdot \cos 2 x + C .$

#### Explanation:

Suppose that, $I = \int \left({x}^{2} + 5\right) \cos 2 x \mathrm{dx} .$

We will use the following Rule of Integration by Parts (IBP) :

$\text{ IBP } : \int u \cdot v \mathrm{dx} = u \int v \mathrm{dx} - \int \left\{\frac{\mathrm{du}}{\mathrm{dx}} \int v \mathrm{dx}\right\} \mathrm{dx} .$

We take, $u = {x}^{2} + 5 , \mathmr{and} , v = \cos 2 x .$

$\therefore \frac{\mathrm{du}}{\mathrm{dx}} = 2 x , \mathmr{and} , \int v \mathrm{dx} = \frac{\sin 2 x}{2.}$

$\therefore I = \frac{{x}^{2} + 5}{2} \cdot \sin 2 x - \int \left\{\left(2 x\right) \frac{\sin 2 x}{2}\right\} \mathrm{dx} ,$

$= \frac{1}{2} \left({x}^{2} + 5\right) \sin 2 x - J , \text{... ...(1), where, } J = \int x \sin 2 x \mathrm{dx} .$

To evaluate $J ,$ we reuse IBP, this time, with,

$u = x , v = \sin 2 x \Rightarrow \frac{\mathrm{du}}{\mathrm{dx}} = 1 , \int v \mathrm{dx} = - \frac{1}{2} \cdot \cos 2 x .$

$\therefore J = - \frac{x}{2} \cdot \cos 2 x - \int \left\{\left(1\right) \left(- \frac{1}{2} \cdot \cos 2 x\right)\right\} \mathrm{dx} ,$

$= - \frac{x}{2} \cdot \cos 2 x + \frac{1}{2} \int \cos 2 x \mathrm{dx} ,$

$= - \frac{x}{2} \cdot \cos 2 x + \frac{1}{2} \cdot \frac{1}{2} \sin 2 x ,$

$\therefore J = - \frac{x}{2} \cdot \cos 2 x + \frac{1}{4} \cdot \sin 2 x \ldots \ldots \ldots \left(2\right) .$

From $\left(1\right) , \mathmr{and} , \left(2\right) ,$ we have,

$I = \frac{1}{2} \left({x}^{2} + 5\right) \sin 2 x + \frac{x}{2} \cdot \cos 2 x - \frac{1}{4} \cdot \sin 2 x + C , \mathmr{and} ,$

$I = \frac{1}{4} \left(2 {x}^{2} + 9\right) \sin 2 x + \frac{x}{2} \cdot \cos 2 x + C .$

Enjoy Maths.!