Question

Evaluate the following definite integral:

\[\int_0^1 \frac{1}{\sqrt{\left( x - 1 \right)\left( 2 - x \right)}}dx\]

Answer 1

Let I =

\[\int_1^2 \frac{1}{\sqrt{\left( x - 1 \right)\left( 2 - x \right)}}dx\]

Put

\[x = \cos^2 \theta + 2 \sin^2 \theta\]

`thereforedx=2costheta(-sintheta)dtheta+4sinthetacostheta d theta=2sinthetacostheta d theta`

Also,

\[x = \cos^2 \theta + 2 \sin^2 \theta\]
\[ \Rightarrow x = 1 + \sin^2 \theta\]
\[ \Rightarrow \sin\theta = \sqrt{x - 1}\]

When `xrarr1, sinthetararr0" or "thetararr0`

When \[x \to 2, \sin\theta \to 1\text{ or }\theta \to \frac{\pi}{2}\]

`therefore I = `\[\int_1^2 \frac{1}{\sqrt{\left( x - 1 \right)\left( 2 - x \right)}}dx\]

\[\Rightarrow I = \int_0^\frac{\pi}{2} \frac{2\sin\theta\cos\theta d\theta}{\sqrt{\left( \cos^2 \theta + 2 \sin^2 \theta - 1 \right)\left( 2 - \cos^2 \theta - 2 \sin^2 \theta \right)}}\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{2\sin\theta\cos\theta d\theta}{\sqrt{\sin^2 \theta \cos^2 \theta}} ...................\left( \sin^2 \theta + \cos^2 \theta = 1 \right)\]
\[ \Rightarrow I = \int_0^\frac{\pi}{2} \frac{2\sin\ theta\cos\theta  d\theta}{\sin\theta\cos\theta}\]
\[ \Rightarrow I = 2 \int_0^\frac{\pi}{2} d\theta\]
\[ \Rightarrow I = 2\theta |_0^\frac{\pi}{2}\]

\[\Rightarrow I = 2\left( \frac{\pi}{2} - 0 \right) = \pi\]