#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.25 Question 27 Maths Textbook Solution.

Answer: $-\left[\sqrt{1-x^{2}} \cos ^{-1} x+x\right]+c$

Given: $\int \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x$

Hint: Use integration by parts

Solution:

$I=\int \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}} d x \Rightarrow-\frac{1}{2} \int \frac{-2 x}{\sqrt{1-x^{2}}} \cos ^{-1} x d x$

[Using integration by parts]

\begin{aligned} &=\frac{-1}{2}\left[\cos ^{-1} x \times \int \frac{-2 x}{\sqrt{1-x^{2}}} d x-\int\left\{\left(\frac{d}{d x} \cos ^{-1} x\right) \int \frac{-2 x}{\sqrt{1-x^{2}}} d x\right\} d x\right] \\ &=\frac{-1}{2}\left[\cos ^{-1} x \cdot 2 \sqrt{1-x^{2}}-\int-\frac{1}{\sqrt{1-x^{2}}} 2 \sqrt{1-x^{2}} d x\right] \\ &=\frac{-1}{2}\left[2 \sqrt{1-x^{2}} \cos ^{-1} x+2 x\right]+c \\ &=-\left[\sqrt{1-x^{2}} \cos ^{-1} x+x\right]+c \end{aligned}

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