# X.COS²X dx ka integration ?

Apr 25, 2018

$I = \frac{1}{4} {x}^{2} + \frac{1}{4} x \sin \left(2 x\right) + \frac{1}{8} \cos \left(2 x\right) + C$

#### Explanation:

We want to solve

$I = \int x {\cos}^{2} \left(x\right) \mathrm{dx}$

Using the trig identity color(red)(cos^2(x)=1/2+1/2cos(2x)

$I = \int x \left(\frac{1}{2} + \frac{1}{2} \cos \left(2 x\right)\right) \mathrm{dx}$

$\textcolor{w h i t e}{I} = \frac{1}{2} \int x \mathrm{dx} + \int x \cos \frac{2 x}{2} \mathrm{dx}$

The first integral is trivial, for the second apply IBP

$\textcolor{b l u e}{\int u \mathrm{dv} = u v - \int v \mathrm{du}}$

Let color(blue)(u=x=>du=dx

And color(blue)(dv=cos(2x)/2dx=>v=sin(2x)/4

${I}_{1} = \int x \cos \frac{2 x}{2} \mathrm{dx}$

$\textcolor{w h i t e}{{I}_{1}} = x \sin \frac{2 x}{4} - \int \sin \frac{2 x}{4} \mathrm{dx}$

$\textcolor{w h i t e}{{I}_{1}} = x \sin \frac{2 x}{4} + \cos \frac{2 x}{8} + {C}_{1}$

Thus

$I = \frac{1}{4} {x}^{2} + \frac{1}{4} x \sin \left(2 x\right) + \frac{1}{8} \cos \left(2 x\right) + C$

Apr 25, 2018

$I = \frac{1}{8} \left[2 {x}^{2} + 2 \sin 2 x + \cos 2 x\right] + C$

#### Explanation:

Here,

$I = \int x {\cos}^{2} x \mathrm{dx}$

$= \int x \left(\frac{1 + \cos 2 x}{2}\right) \mathrm{dx}$

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

$I = \frac{1}{2} \cdot {x}^{2} / 2 + \frac{1}{2} {I}_{1.} . . \to \left(A\right)$

Where, ${I}_{1} = \int x \cos 2 x \mathrm{dx}$

$\text{Using"color(blue)" Integration by Parts:}$

$\int \left(u v\right) \mathrm{dx} = u \int v \mathrm{dx} - \int \left(u ' \int v \mathrm{dx}\right) \mathrm{dx}$

Let, $u = x \mathmr{and} v = \cos 2 x \implies u ' = 1 \mathmr{and} \int v \mathrm{dx} = \frac{\sin 2 x}{2}$

So,

${I}_{1} = x \cdot \frac{\sin 2 x}{2} - \int 1 \cdot \frac{\sin 2 x}{2} \mathrm{dx}$

$= \frac{x}{2} \sin 2 x - \frac{1}{2} \int \sin 2 x \mathrm{dx}$

$= \frac{x}{2} \sin 2 x - \frac{1}{2} \left(\frac{- \cos 2 x}{2}\right) + c$

$\therefore {I}_{1} = \frac{x}{2} \sin 2 x + \frac{\cos 2 x}{4} + c$

Hence, from $\left(A\right)$

$I = \frac{1}{2} \cdot {x}^{2} / 2 + \frac{1}{2} \left[\frac{x}{2} \sin 2 x + \frac{\cos 2 x}{4} + c\right]$

$= {x}^{2} / 4 + \frac{x}{4} \sin 2 x + \frac{1}{8} \cos 2 x + \frac{c}{2}$

$I = \frac{1}{8} \left[2 {x}^{2} + 2 \sin 2 x + \cos 2 x\right] + C , w h e r e . C = \frac{c}{2}$