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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS5 Marks
Question
Evaluate the following:
$\int\limits^{\frac{1}{2}}_0\frac{\text{dx}}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}$
Hint: let $\text{x}=\sin\theta$
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Answer

Let $\text{I}=\int\limits^{\frac{1}{2}}_0\frac{\text{dx}}{(1+\text{x}^2)\sqrt{1-\text{x}^2}}$
Put $\text{x}=\sin\theta$
$\Rightarrow\ \text{dx}=\cos\theta\text{ d}\theta$
As $\text{x}\rightarrow0,$ then $\theta\rightarrow0$
and $\text{x}\rightarrow\frac{1}{2},$ then $\theta\rightarrow\frac{\pi}{6}$
$\therefore\ \text{I}=\int\limits^{\frac{\pi}{6}}_0\frac{\cos\theta}{(1+\sin^2\theta)\cos\theta}\text{d}\theta$ $=\int\limits^{\frac{\pi}{6}}_0\frac{1}{1+\sin^2\theta}\text{d}\theta$
$=\int\limits^{\frac{\pi}{6}}_0\frac{1}{\cos^2\theta(\sec^2\theta+\tan^2\theta)}\text{d}\theta$
$=\int\limits^{\frac{\pi}{6}}_0\frac{\sec^2\theta}{\sec^2\theta+\tan^2\theta}\text{d}\theta$
$=\int\limits^{\frac{\pi}{6}}_0\frac{\sec^2\theta}{1+\tan^2\theta+\tan^2\theta}\text{d}\theta$
$=\int\limits^{\frac{\pi}{6}}_0\frac{\sec^2\theta}{1+2\tan^2\theta}\text{d}\theta$
Again, put $\tan\theta=\text{t}$
$\Rightarrow\ \sec^2\theta\text{ d}\theta=\text{dt}$
As $\theta\rightarrow0,$ then $\text{t}\rightarrow0$
and $\theta\rightarrow\frac{\pi}{6},$ then $\text{t}\rightarrow\frac{1}{\sqrt{3}}$
$\therefore\ \text{I}=\int\limits^{\frac{1}{\sqrt{3}}}_0\frac{\text{dt}}{1+2\text{t}^2}=\frac{1}{2}\int\limits^{\frac{1}{\sqrt{3}}}_0$ $\frac{\text{dt}}{\Big(\frac{1}{\sqrt{2}}\Big)^2+\text{t}^2}$
$=\frac{1}{2}\cdot\frac{1}{\frac{1}{\sqrt{2}}}\Bigg[\tan^{-1}\frac{\text{t}}{\frac{1}{\sqrt{2}}}\Bigg]^{\frac{1}{\sqrt{3}}}_0$ $=\frac{1}{\sqrt{2}}\Big[\tan^{-1}(\sqrt{2}\text{t})\Big]^{\frac{1}{\sqrt{3}}}_0$
$=\frac{1}{\sqrt{2}}\bigg[\tan^{-1}\sqrt{\frac{2}{3}}-0\bigg]$ $=\frac{1}{\sqrt{2}}\tan^{-1}\Big(\sqrt{\frac{2}{3}}\Big)$

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