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Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 110 Maths Textbook Solution.

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Answer: \frac{\sqrt{x} \sqrt{1-x}}{2}-\frac{\sin ^{-1} \sqrt{x}(1-2 x)}{2}+C

Hint: to solve the question we have to use ILATE method

Given: \int \sin ^{-1} \sqrt{x} d x


I=\int \sin ^{-1} \sqrt{x} d x

\text { Let } \sqrt{x}=\sin t

x=\sin ^{2} t \ldots \text { squaring on both sides }

d x=\sin t \cos t d t

I=\int \sin ^{-1}(\sin t) \sin t \cos t d t

I=\frac{1}{2} \int t \sin 2 t d t

\left[\int(u . v) d x=u \int v d x-\int\left[\frac{d}{d x} u \cdot \int u d x\right]\right] d x

I=\frac{1}{2}\left\{t \frac{\cos 2 t}{2}+\int \frac{\cos 2 t}{2} d t\right\}

I=\frac{1}{2}\left\{\frac{-t \cos 2 t}{2}+\frac{\sin 2 t}{4}\right\}+C

\sqrt{x}=\sin t

t=\sin \sqrt{x}

\cos t=\sqrt{1-\sin ^{2} t}

\cos t=\sqrt{1-x}

I=-\sin ^{-1} \frac{\sqrt{x}(1-2 x)}{2}+\frac{2 \sqrt{x} \sqrt{1-x}}{4}+C

I=\frac{\sqrt{x} \sqrt{1-x}}{2}-\frac{\sin ^{-1} \sqrt{x}(1-2 x)}{2}+C

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