#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise  Question 91 Maths Textbook Solution.

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

Hint:

You must know about integration of cos2x.

Given:

$\int(1+x)^{2} \cos 2 x d x$

Solution:

$\int(1+x)^{2} \cos 2 x d x$

$\int \cos 2 x d x+\int x^{2} \cos 2 x d x$

$\frac{\sin 2 x}{2}+\left[\frac{x^{2} \sin 2 x}{2}-\int 2 x\left(\frac{\sin 2 x}{2}\right) d x\right.$                            $\left[1 . I 1 d x=I \int I I d x-\left[\left(\frac{d}{d x}\right) I \int I I d x\right]\right.$

$\frac{\sin 2 x}{2}+\left[\frac{x^{2} \sin 2 x}{2}-\left[x \frac{-\cos 2 x}{2}+-\frac{1}{2} \int \cos 2 x d x\right.\right.$

$\frac{\sin 2 x}{2}+\frac{x^{2} \sin 2 x}{2}+\frac{x \cos 2 x}{2}-\frac{1}{2} \int \cos 2 x d x$

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

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

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