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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.

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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