Question
Evaluate the following definite integrals:
$\int\limits_{0}^{1}\frac{1}{\sqrt{(\text{x}-1)(2-\text{x})}}\text{ dx}$
$\int\limits_{0}^{1}\frac{1}{\sqrt{(\text{x}-1)(2-\text{x})}}\text{ dx}$
✓
Answer
Let $\text{I}=\int_{0}^\limits{1}\frac{1}{\sqrt{(\text{x}-1)(2-\text{x})}}\text{ dx}$
Put $\text{x}=\cos^2\theta+2\sin^2\theta$
$\therefore\ \text{dx}=2\cos\theta(-\sin\theta)\text{d}\theta+4\sin\theta\cos\theta\text{ d}\theta$
$=2\sin\theta\cos\theta\text{ d}\theta$
Also, $\text{x}=\cos^2\theta+2\sin^2\theta$
$\Rightarrow\text{x}=1+\sin^2\theta$
$\Rightarrow\sin\theta=\sqrt{\text{x}-1}$
When $\text{x}\rightarrow1,\sin\theta\rightarrow0$ or $\theta\rightarrow0$
When $\text{x}\rightarrow2,\sin\theta\rightarrow1$ or $\theta\rightarrow\frac{\pi}{2}$
$\therefore\ \text{I}=\int\limits^2_1\frac{1}{\sqrt{(\text{x}-1)(2-\text{x})}}\text{ dx}$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{2\sin\theta\cos\theta\text{ d}\theta}{\sqrt{(\cos^2\theta+2\sin^2\theta-1)(2-\cos^2\theta-2\sin^2\theta)}}$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{2\sin\theta\cos\theta\text{ d}\theta}{\sqrt{\sin^2\theta\cos^2\theta}}$ $\big(\sin^2\theta+\cos^2\theta=1\big)$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{2\sin\theta\cos\theta\text{ d}\theta}{\sin\theta\cos\theta}$
$\Rightarrow\text{I}=2\int\limits^\frac{\pi}{2}_0\text{d}\theta$
$\Rightarrow\text{I}=2\big[\theta\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}-0\Big)$
$\Rightarrow\text{I}=\pi$
Put $\text{x}=\cos^2\theta+2\sin^2\theta$
$\therefore\ \text{dx}=2\cos\theta(-\sin\theta)\text{d}\theta+4\sin\theta\cos\theta\text{ d}\theta$
$=2\sin\theta\cos\theta\text{ d}\theta$
Also, $\text{x}=\cos^2\theta+2\sin^2\theta$
$\Rightarrow\text{x}=1+\sin^2\theta$
$\Rightarrow\sin\theta=\sqrt{\text{x}-1}$
When $\text{x}\rightarrow1,\sin\theta\rightarrow0$ or $\theta\rightarrow0$
When $\text{x}\rightarrow2,\sin\theta\rightarrow1$ or $\theta\rightarrow\frac{\pi}{2}$
$\therefore\ \text{I}=\int\limits^2_1\frac{1}{\sqrt{(\text{x}-1)(2-\text{x})}}\text{ dx}$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{2\sin\theta\cos\theta\text{ d}\theta}{\sqrt{(\cos^2\theta+2\sin^2\theta-1)(2-\cos^2\theta-2\sin^2\theta)}}$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{2\sin\theta\cos\theta\text{ d}\theta}{\sqrt{\sin^2\theta\cos^2\theta}}$ $\big(\sin^2\theta+\cos^2\theta=1\big)$
$\Rightarrow\text{I}=\int\limits^\frac{\pi}{2}_0\frac{2\sin\theta\cos\theta\text{ d}\theta}{\sin\theta\cos\theta}$
$\Rightarrow\text{I}=2\int\limits^\frac{\pi}{2}_0\text{d}\theta$
$\Rightarrow\text{I}=2\big[\theta\big]^{\frac{\pi}{2}}_0$
$\Rightarrow\text{I}=2\Big(\frac{\pi}{2}-0\Big)$
$\Rightarrow\text{I}=\pi$
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