Evaluate: _0^ 4 x 4- ^2 x d x - Vidyadip

Question

Evaluate: $\int_0^{\frac{\pi}{4}} \frac{\cos x}{4-\sin ^2 x} d x$

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Answer

Let $I =\int_0^{\frac{\pi}{4}} \frac{\cos x}{4-\sin ^2 x} d x$
Put $\sin x=t$
$\therefore \cos x d x=d t$
When $x =0, t =0$ and when $x =\frac{\pi}{4}, t \frac{1}{\sqrt{2}}$
$ \therefore I=\int_0^{\frac{1}{\sqrt{2}}} \frac{ dt }{4- t ^2}$
$=\int_0^{\frac{1}{\sqrt{2}}} \frac{ dt }{2^2- t ^2}$
$=\left[\frac{1}{2 \times 2} \log \left|\frac{2+ t }{2- t }\right|\right]_0^{\frac{1}{\sqrt{2}}}$
$=\frac{1}{4}\left[\log \left|\frac{2+\frac{1}{\sqrt{2}}}{2-\frac{1}{\sqrt{2}}}\right|-\log 1\right.$
$=\frac{1}{4}\left[\log \left|\frac{2 \sqrt{2}+1}{2 \sqrt{2}-1}\right|-0\right]$
$\therefore I =\frac{1}{4} \log \left(\frac{2 \sqrt{2}+1}{2 \sqrt{2}-1}\right) $

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