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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.

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

Posted by

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