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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.

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


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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