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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 51

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

Hints: You must know about the integral rule of trigonometric functions.

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


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

Let x=\sin(u)

\begin{aligned} &d x=\cos \mathrm{u} \mathrm{d} \mathrm{u} \\ &\int \frac{1}{\sqrt{1-x^{2}}} d x=\int \frac{1}{1-\sin ^{2} u} \cos u d u \\ &\sin ^{2} u+\cos ^{2} u=1 \\ &\cos ^{2} u=1-\sin ^{2} u \end{aligned}

\begin{aligned} &=\int \frac{\cos (u)}{\sqrt{\cos ^{2}(u)}} d u \\ &=\int \frac{\cos u}{\cos u} d u \\ &=\int 1 d u \\ &=u+c\quad \quad \quad \quad \quad \quad \left [ \int x^{n}dx =\frac{x^{n+1}}{n+1}\right ] \end{aligned}
Where x= \sin u andu=\sin^{-1}x+c




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