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#### Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 35

Answer: $\frac{1}{2}x\sqrt{4-x^{2}}+2\sin^{-1}\left ( \frac{x}{2} \right )+c$

Hints: You must know about the integral rule of x functions

Given : $\int \sqrt{4-x^{2}}dx$

Solution: $\int \sqrt{4-x^{2}}dx$

\begin{aligned} &x=2 \sin \theta \\ &d x=2 \cos \theta d \theta \\ &\sqrt{4-x^{2}}=\sqrt{4-4 \sin ^{2}} \theta=2 \cos \theta \\ &\int \sqrt{4-x^{2}} d x=\int 2 \cos \theta \cdot 2 \cos \theta d \theta \\ &=4 \int \cos ^{2} \theta d \theta \end{aligned}

We know that $\frac{\cos 2\theta+1}{2}=\cos^{2}\theta$

So,

$4\int \cos ^{2}\theta d\theta =2\int \left ( \cos 2\theta +1 \right )d\theta$

$=2\int \cos 2\theta d\theta +2\int 1d \theta$                                    $\begin{bmatrix} \int \cos axdx=\frac{\sin ax}{a}\\\\\int x^{n }dx=\frac{x^{n+1}}{n+1} \end{bmatrix}$

$=2\sin \theta \cos \theta+2\theta +c$                                       $\left [ \sin 2 x=2\sin x\cos x \right ]$

Now put the values of $\theta,\sin \theta,\cos \theta$

We know $x=2\sin \theta$

$\theta=\sin^{-1}x$

$x=2\sin \theta$

Squaring on both sides

\begin{aligned} &x^{2}=4 \sin ^{2} \theta \\ &x^{2}=4\left(1-\cos ^{2} \theta\right) \\ &4-x^{2}=4 \cos ^{2} \theta \\ &\cos \theta=\sqrt{\frac{4-x^{2}}{2}} \\ &\quad=2 \cdot \frac{x}{2} \cdot \frac{\sqrt{4-x^{2}}}{2}+2 \sin ^{-1}\left(\frac{x}{2}\right) \\ &\quad=\frac{x \cdot \sqrt{4-x^{2}}}{2}+2 \sin ^{-1}\left(\frac{x}{2}\right)+c \end{aligned}