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

Answers (1)


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

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

Let   I=\int \sin ^{3}(2 x+1) d x

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

So the answer is,

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

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