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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 95 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 95 Maths Textbook Solution.

Answers (1)

Answer:

-\frac{3}{4} x \cos x+\frac{3}{4} \sin x+\frac{1}{12} x \cos 3 x-\frac{\sin 3 x}{36}+c

Hint:

You must know about formula of sin3xdx.

Given:

\int x \sin ^{3} x d x

Solution:

\text { now } \sin 3 x=3 \sin x-4 \sin ^{3} x

   4 \sin ^{3} x=3 \sin x-\sin 3 x

\sin ^{3} x=\frac{1}{4}(3 \sin x-\sin 3 x)

\frac{1}{4} \int 3 x \sin x d x-\frac{1}{4} \int x \sin 3 x d x

\frac{3}{4} \int x \sin x d x-\frac{1}{4} \int x \sin 3 x d x

         I    II                            I    II

\frac{3}{4} \int x(-\cos x)+\int 1 .(\cos x) d x-\frac{1}{4}\left[x\left(-\frac{\cos 3 x}{3}+\int 1\left(\frac{\cos 3 x}{3}\right)\right)\right.

\frac{3}{4}[-x \cos x+\sin x]-\frac{1}{4}\left[\frac{x(-\cos 3 x)}{3}\right]-\frac{1}{12} \times \frac{\sin 3 x}{3}

=-\frac{3}{4} x \cos x+\frac{3}{4} \sin x+\frac{1}{12} x \cos 3 x-\frac{\sin 3 x}{36}+c

 

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