Question

Evaluate the following definite integrals:

${\int }_0^1\frac{1}{\sqrt{\left(x-1\right)\left(2-x\right)}}dx$ [NCERT EXEMPLAR]

Answer 1

Let I = ${\int }_1^2\frac{1}{\sqrt{\left(x-1\right)\left(2-x\right)}}dx$

Put $x={\cos }^2\theta +2{\sin }^2\theta$

$\therefore dx=2\cos \theta \left(-\sin \theta \right)d\theta +4\sin \theta \cos \theta d\theta =2\sin \theta \cos \theta d\theta$

Also,

$$\begin{gathered} x={\cos }^2\theta +2{\sin }^2\theta \\ \Rightarrow x=1+{\sin }^2\theta \\ \Rightarrow \sin \theta =\sqrt{x-1} \end{gathered}$$

When $x\to 1,\sin \theta \to 0\mathrm{or}\theta \to 0$

When $x\to 2,\sin \theta \to 1\mathrm{or}\theta \to \frac{\mathrm{\pi }}{2}$

∴ I = ${\int }_1^2\frac{1}{\sqrt{\left(x-1\right)\left(2-x\right)}}dx$
$$\begin{gathered} \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2\sin \theta \cos \theta d\theta }{\sqrt{\left({\cos }^2\theta +2{\sin }^2\theta -1\right)\left(2-{\cos }^2\theta -2{\sin }^2\theta \right)}} \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2\sin \theta \cos \theta d\theta }{\sqrt{{\sin }^2\theta {\cos }^2\theta }}\left({\sin }^2\theta +{\cos }^2\theta =1\right) \\ \Rightarrow I={\int }_0^{\frac{\mathrm{\pi }}{2}}\frac{2\sin \theta \cos \theta d\theta }{\sin \theta \cos \theta } \\ \Rightarrow I=2{\int }_0^{\frac{\mathrm{\pi }}{2}}d\theta \\ \Rightarrow I=2\theta {|}_0^{\frac{\mathrm{\pi }}{2}} \end{gathered}$$
$\Rightarrow I=2\left(\frac{\mathrm{\pi }}{2}-0\right)=\mathrm{\pi }$