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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 15 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 15 Maths Textbook Solution.

Answers (1)

Answer:

\frac{\left(\sin ^{-1} x\right)^{4}}{4}+c

Given:

\int \frac{\left(\sin ^{-1} x\right)^{3}}{\sqrt{1-x^{2}}} d x

Hint:

We must know about substitution formula to integrate.

Solution:   

\int \frac{\left(\sin ^{-1} x\right)^{3}}{\sqrt{1-x^{2}}} d x

Let \mathrm{u}=\sin ^{-1} \mathrm{x}

Differentiate it with respect to x

\frac{d u}{d x}=\frac{1}{\sqrt{1-x^{2}}}

d u=\frac{1}{\sqrt{1-x^{2}}} d x

now, \int u^{3} d u                                (p u t u \& d u)

=\frac{u^{4}}{4}+c

\frac{\left(\sin ^{-1} x\right)^{4}}{4}+c

Posted by

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