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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.

Answers (1)

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

Solution:

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

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

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

$d x=\sin t \cos t d t$

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

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

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

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

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

$\sqrt{x}=\sin t$

$t=\sin \sqrt{x}$

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

$\cos t=\sqrt{1-x}$

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

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

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