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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 55 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 55 Maths Textbook Solution.

Answers (1)

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

Hint:

Given:\int x\sin ^{3}xdx

Solution: \therefore \sin 3x=3\sin x-4\sin ^{3}x 

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

                =\frac{1}{4}\int 3x\sin x-x\sin 3xdx

                =\frac{3}{4}\int x\sin\: x-\frac{1}{4}\int x\sin 3xdx

Take the first function as x and second function as sinx and sin 3x

            \begin{aligned} &=\frac{3}{4}\left[x \int \sin x-\int\left[\frac{d x}{d x} \int \sin x d x\right] d x\right]-\frac{1}{4}\left[x \int(\sin 3 x d x)-\int\left[\frac{d x}{d x} \int \sin 3 x d x\right] d x\right] \\ &=\frac{3}{4}\left(-x \cos x+\int \cos x d x\right)-\frac{1}{4}\left[\frac{-x \cos 3 x}{3}+\int \frac{\cos 3 x}{3} d x\right] \\ &=\frac{3}{4}(-x \cos x+\sin x)-\frac{1}{4}\left[\frac{-x \cos 3 x}{3}+\frac{\sin 3 x}{9}\right]+c \\ &=\frac{-3 x \cos x}{4}+\frac{3 \sin x}{4}+\frac{x \cos 3 x}{12}-\frac{\sin 3 x}{36}+c \end{aligned}

 

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