#### Please Solve R.D.Sharma class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 59 Maths textbook Solution.

Answer: $\pi$

Hint: Use indefinite formula then put the limit to solve this integral

Given: $\int_{1}^{2}\frac{1}{\sqrt{\left ( x-1 \right )\left ( 2-x \right )}}dx$

Solution:

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

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

$=\int_{1}^{2} \frac{1}{\sqrt{-\left(x^{2}-2 x \cdot \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+2\right)}} d x \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[a^{2}+b^{2}-2 a b=(a-b)^{2}\right]$

$=\int_{1}^{2} \frac{1}{\sqrt{-\left(\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}+2\right)}} d x$

\begin{aligned} &=\int_{1}^{2} \frac{1}{\sqrt{-\left(\left(x-\frac{3}{2}\right)^{2}-\left(\frac{9-8}{4}\right)\right)}} d x \\ &=\int_{1}^{2} \frac{1}{\sqrt{-\left(\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{4}\right)\right)}} d x \\ &=\int_{1}^{2} \frac{1}{\sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}}} d x \end{aligned}

Putting $\left ( x-\frac{3}{2} \right )=t\Rightarrow dx=dt$

When $x=1$ then $t=1-\frac{3}{2}=\frac{2-3}{2}=\frac{-1}{2}$

And When $x=2$  then $t=2-\frac{3}{2}=\frac{4-3}{2}=\frac{1}{2}$

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

$=\left[\sin ^{-1}\left(\frac{t}{\frac{1}{2}}\right)\right]_{\frac{-1}{2}}^{\frac{1}{2}}$                                                            $\left[\int \frac{1}{a^{2}-x^{2}} d x=\sin ^{-1} \frac{x}{a}\right]$

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

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

$=\sin ^{-1}(1)-\sin ^{-1}(-1)$

$=\sin ^{-1}(1)+\sin ^{-1}(1) \\$

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

$=2\times \frac{\pi }{2}$

$=\pi$