#### Explain Solution R.D.Sharma Class 12 Chapter 19 Definite Integrals  Exercise 19.1 Question 42 Maths Textbook Solution.

$\frac{\pi }{3}$

Hint: Use indefinite formula and put the limits to solve this integral

Given:

$\int_{0}^{2}\frac{1}{\sqrt{3+2x-x^{2}}}dx$

Solution:

$\int_{0}^{2}\frac{1}{\sqrt{3+2x-x^{2}}}dx=\int_{0}^{2}\frac{1}{\sqrt-{\left ( x^{2}-2x-3 \right )}}dx$

$=\int_{0}^{2} \frac{1}{\sqrt{-\left\{x^{2}-2 x \cdot 1+1-1-3\right\}}} d x$

$=\int_{0}^{2} \frac{1}{\sqrt{-\left\{(x-1)^{2}-1-3\right\}}} d x$

$\left [ \because \left ( a-b \right )^{2}=a^{2}-2ab+b^{2} \right ]$

$=\int_{0}^{2} \frac{1}{\sqrt{-\left\{(x-1)^{2}-4\right\}}} d x$

$=\int_{0}^{2} \frac{1}{\sqrt{4-(x-1)^{2}}} d x$

$=\int_{0}^{2} \frac{1}{\sqrt{2^{2}-(x-1)^{2}}} d x$

$=\left[\sin ^{-1}\left(\frac{x-1}{2}\right)\right]_{0}^{2}$

$\left [ \because \int \frac{1}{\sqrt{a^{2}-x^{2}}}dx=\sin ^{-1}\frac{x}{a} \right ]$

$=\left[\sin ^{-1}\left(\frac{2-1}{2}\right)-\sin ^{-1}\left(\frac{0-1}{2}\right)\right]$

$=\left[\sin ^{-1}\left(\frac{1}{2}\right)-\sin ^{-1}\left(\frac{-1}{2}\right)\right]$

$\left [ \because \sin ^{-1}\left ( -\theta \right )=-\sin \theta \right ]$

$=\left[\sin ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(\frac{1}{2}\right)\right]$

$=2 \sin ^{-1}\left(\frac{1}{2}\right)$

$\left [ \because \sin ^{-1}\frac{1}{2}=\frac{\pi }{6} \right ]$

$=2.\frac{\pi }{6}$

$=\frac{\pi }{3}$