Question
Evaluate: $\int_0^{\frac{1}{2}} \frac{1}{\left(1-2 x^2\right) \sqrt{1-x^2}} d x$
✓
Answer
Let $I =\int_0^{\frac{1}{2}} \frac{1}{\left(1-2 x^2\right) \sqrt{1-x^2}} d x$
Put $x=\sin \theta$
$
\therefore dx =\cos \theta d \theta
$
When $x=0, \theta=\sin ^{-1} 0=0$ and
When $x=\frac{1}{2}, \theta=\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$
$
\therefore I=\int_0^{\frac{\pi}{6}} \frac{1}{\left(1-2 \sin ^2 \theta\right)\left(\sqrt{1-\sin ^2 \theta}\right)} \cos \theta d \theta
$
$=\int_0^{\frac{\pi}{6}} \frac{1}{(\cos 2 \theta)(\cos \theta)} \cos \theta d \theta$
$=\int_0^{\frac{\pi}{6}} \frac{1}{\cos 2 \theta} \cdot d \theta$
$=\int_0^{\frac{\pi}{6}} \sec 2 \theta d \theta$
$=\frac{1}{2}[\log |\sec 2 \theta+\tan 2 \theta|]_0^{\frac{\pi}{6}}$
$=\frac{1}{2}\left[\log \left|\sec 2\left(\frac{\pi}{6}\right)+\tan 2\left(\frac{\pi}{6}\right)\right|\right]-\frac{1}{2}[\log |\sec 2(0)+\tan 2(0)|]$
$=\frac{1}{2}\left[\log \left|\sec \left(\frac{\pi}{3}\right)+\tan \left(\frac{\pi}{3}\right)\right|\right]-\frac{1}{2}[\log |\sec (0)+\tan (0)|]$
$=\frac{1}{2}[\log |2+\sqrt{3}|]-\frac{1}{2} \log |1|$
$\therefore 1=\frac{1}{2} \log |2+\sqrt{3}|$
Put $x=\sin \theta$
$
\therefore dx =\cos \theta d \theta
$
When $x=0, \theta=\sin ^{-1} 0=0$ and
When $x=\frac{1}{2}, \theta=\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$
$
\therefore I=\int_0^{\frac{\pi}{6}} \frac{1}{\left(1-2 \sin ^2 \theta\right)\left(\sqrt{1-\sin ^2 \theta}\right)} \cos \theta d \theta
$
$=\int_0^{\frac{\pi}{6}} \frac{1}{(\cos 2 \theta)(\cos \theta)} \cos \theta d \theta$
$=\int_0^{\frac{\pi}{6}} \frac{1}{\cos 2 \theta} \cdot d \theta$
$=\int_0^{\frac{\pi}{6}} \sec 2 \theta d \theta$
$=\frac{1}{2}[\log |\sec 2 \theta+\tan 2 \theta|]_0^{\frac{\pi}{6}}$
$=\frac{1}{2}\left[\log \left|\sec 2\left(\frac{\pi}{6}\right)+\tan 2\left(\frac{\pi}{6}\right)\right|\right]-\frac{1}{2}[\log |\sec 2(0)+\tan 2(0)|]$
$=\frac{1}{2}\left[\log \left|\sec \left(\frac{\pi}{3}\right)+\tan \left(\frac{\pi}{3}\right)\right|\right]-\frac{1}{2}[\log |\sec (0)+\tan (0)|]$
$=\frac{1}{2}[\log |2+\sqrt{3}|]-\frac{1}{2} \log |1|$
$\therefore 1=\frac{1}{2} \log |2+\sqrt{3}|$
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