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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 19 Maths Textbook Solution.

Answers (1)

Answer: \frac{-x \cos 2 x}{4}+\frac{\sin 2 x}{8}+c

Hint: We know that,

        Sin 2x = 2 sinxcosx, it can be written as \int x \sin x \cos x d x=\frac{1}{2} \int x \sin 2 x d xand consider first

function as x and second function as sin2x

Given: Let I=\int x \sin x \cos x d x

Solution: \frac{1}{2} \int x \sin 2 x d x

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

Again by integrating second term

               =\frac{1}{2}\left(\frac{-x \cos 2 x}{2}+\frac{\sin 2 x}{4}\right)+c

By further simplification,

=\frac{-x \cos 2 x}{4}+\frac{\sin 2 x}{8}+c

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