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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 9 Maths Textbook Solution.

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Answer: \frac{-1}{2} x \cos 2 x+\frac{1}{4} \sin 2 x+c

Hint: Use integrate on by parts

         \int u v d x=u \int v d x-\int \frac{d}{d x} u \int v d x

Given: I=\int x \sin 2 x d x

Solution: I=\int x \sin 2 x d x

                \Rightarrow d(\cos 2 x)=-2 \sin 2 x d x

We can integrate by parts in this way

               \begin{aligned} &=\int x \sin 2 x d x \\ &=\frac{-1}{2} \int x d(\cos 2 x) \\ &=\frac{-1}{2} x \cos 2 x+\frac{1}{2} \int \cos 2 x d x \\ &=\frac{-1}{2} x \cos 2 x+\frac{1}{4} \sin 2 x+c \end{aligned}


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