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Definite Integration — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDefinite Integration4 Marks
Question
Evaluate : $\int_0^{1 / 2} \frac{1}{\left(1-2 x^2\right) \sqrt{1-x^2}} \cdot d x$
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Answer

Let $\quad I=\int_0^{1 / 2} \frac{1}{\left(1-2 x^2\right) \sqrt{1-x^2}} \cdot d x$
put $x=\sin \theta \quad \therefore \quad 1 \cdot d x=\cos \theta \cdot d \theta$
As $\quad x$ varies from 0 to $\frac{1}{2}, \theta$ varies from 0 to $\frac{\pi}{6}$
$
\begin{aligned}
& =\int_0^{\pi / 6} \frac{\cos \theta}{\left(1-2 \sin ^2 \theta\right) \sqrt{1-\sin ^2 \theta}} \cdot d \theta=\int_0^{\pi / 6} \frac{\cos \theta}{(\cos 2 \theta) \sqrt{\cos ^2 \theta}} \cdot d \theta \\
& =\int_0^{\pi / 6} \frac{1}{\cos 2 \theta} \cdot d \theta \\
& =\int_0^{\pi / 6} \sec 2 \theta \cdot d \theta \\
& =\left[\log (\sec 2 \theta+\tan 2 \theta) \cdot \frac{1}{2}\right]_0^{\pi / 6} \\
& =\frac{1}{2} \cdot\left[\log \left(\sec 2\left(\frac{\pi}{6}\right)+\tan 2\left(\frac{\pi}{6}\right)-\log (\sec 0+\tan 0)\right]\right. \\
& =\frac{1}{2} \cdot\left[\log \left(\sec \frac{\pi}{3}+\tan \frac{\pi}{3}\right)-\log (1+0)\right] \\
& =\frac{1}{2} \cdot[\log (2+\sqrt{3})-0] \\
& =\frac{1}{2} \log (2+\sqrt{3})
\end{aligned}
$
$
\therefore \quad \int_0^{1 / 2} \frac{1}{\left(1-2 x^2\right) \sqrt{1-x^2}} \cdot d x=\frac{1}{2} \log (2+\sqrt{3})
$

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