#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise 18.6 Question 2

$-\frac{3}{8} \cos (2 x+1)+\frac{1}{24} \cos (6 x+3)+C$
Hint:

Use:  $\sin 3 x=-4 \sin ^{3} x+3 \sin x$
Given:

Let   $I=\int \sin ^{3}(2 x+1) d x$
Solution:

$I=\int \sin ^{3}(2 x+1) d x$
The above stated formula can be written as
$\sin ^{3} x=\frac{3 \sin x-\sin 3 x}{4}$

The equation becomes,

$\sin ^{3} x=\frac{3 \sin x-\sin 3 x}{4}$

On multiplying the above formula to the given question, we get

$\Rightarrow \int \sin ^{3}(2 x+1) d x=\int \frac{3 \sin (2 x+1)-\sin 3(2 x+1)}{4} d x$

We know,

$\int \sin ax=-\frac{1}{a} \cos a x+c$

On substituting the above formula we get ,

$\Rightarrow \frac{3}{4} \int \sin (2 x+1) d x-\frac{1}{4} \int \sin (6 x+3) d x$

On integrating we get ,

$=-\frac{3}{8} \cos (2 x+1)+\frac{1}{24} \cos (6 x+3)+C$

$=-\frac{3}{8} \cos (2 x+1)+\frac{1}{24} \cos (6 x+3)+C$

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