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Please Solve R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 50  Maths Textbook Solution.

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Answer: \frac{\sin 3x}{18}-\frac{x\cos 3x}{6}+\frac{x\cos x}{2}-\frac{\sin x}{2}+c

Hint: \int sin\: xdx=\cos \: x&\int \cos \: xdx=-\sin \: x

Given: \int\: x \sin x\cos 2xdx


            I=\int\: x \sin x\cos 2xdx

            I=\frac{1}{2}\int\: 2x \sin x\cos 2xdx

            =\frac{1}{2}\int\: x2 \sin x\cos 2xdx

            \begin{aligned} &=\frac{1}{2} \int x[\sin (x+2 x)+\sin (x-2 x)] d x \\ &=\frac{1}{2} \int x \sin 3 x d x-\frac{1}{2} \int x \sin x d x \\ &=\frac{1}{2}\left[x \frac{\cos 3 x}{3}+\frac{1}{3} \int \cos 3 x d x\right]-\frac{1}{2}\left[x \cos x+\int \cos x d x\right] \\ &I=\frac{1}{2}\left[-x \frac{\cos 3 x}{3}+\frac{\sin 3 x}{9}\right]-\frac{1}{2}[-x \cos x+\sin x] \\ &=\frac{1}{2}\left[\frac{\sin 3 x}{9}-\frac{x \cos 3 x}{3}+x \cos x-\sin x\right]+c \\ &=\frac{\sin 3 x}{18}-\frac{x \cos 3 x}{6}+\frac{x \cos x}{2}-\frac{\sin x}{2}+c \end{aligned}

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