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

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

Answers (1)

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}

               

Posted by

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