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

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

Answers (1)

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

Solution:

            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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