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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise 18.25 Question 52 Maths Textbook Solution.

Answers (1)

Answer: -\sqrt{1-x^{2}}\sin ^{-1}x+x+c

Hint: Put \sin ^{-1}x=t

Given: \int \frac{x\sin ^{-1}x}{\sqrt{1-x^{2}}}dx

Solution:By letting \sin ^{-1}x=t\Rightarrow x=\sin t\Rightarrow dx=\cos tdt

               \begin{aligned} \therefore \int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x &=\int \frac{\sin t . t}{\sqrt{1-\sin ^{2} t}} \cos t d t \\ &=\int \frac{t \sin t \cos t}{\cos t} d t \\ &=\int t \sin t d t \Rightarrow t(-\cos t)-\int 1(-\cos t) d t \end{aligned}

Applying by parts

                                                 =t\cos t+\sin t+c

                                                  =t\sqrt{1-t}+\sin t+c

                                                   =\sqrt{1-x^{2}}\sin ^{-1}x+x+c

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