Question
Evaluate: $\int_0^{\frac{\pi}{2}} \frac{1}{5+4 \cos x} d x$
✓
Answer
Let $I =\int_0^{\frac{\pi}{2}} \frac{1}{5+4 \cos x} d x$
Put $\tan \left(\frac{x}{2}\right)= t$
$\therefore x =2 \tan ^{-1} t$
$\therefore dx =\frac{2}{1+ t ^2} dt$ and $\cos x =\frac{1- t ^2}{1+ t ^2}$
When $x =0, t =0$ and when $x =\frac{\pi}{2}, t =1$
$\therefore I =\int_0^1 \frac{1}{5+4\left(\frac{1- t ^2}{1+ t ^2}\right)} \times \frac{2}{1+ t ^2} dt$
$=2 \int_0^1 \frac{1}{5+5 t +4-4 t ^2} dt$
$=2 \int_0^1 \frac{1}{9+ t ^2} dt$
$=2 \int_0^1 \frac{1}{ t ^2+3^2} dt$
$=2\left[\frac{1}{3} \tan ^{-1}\left(\frac{ t }{3}\right)\right]_0^1$
$=\frac{2}{3}\left[\tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}(0)\right]$
$=\frac{2}{3} \tan ^{-1}\left(\frac{1}{3}\right)$
Put $\tan \left(\frac{x}{2}\right)= t$
$\therefore x =2 \tan ^{-1} t$
$\therefore dx =\frac{2}{1+ t ^2} dt$ and $\cos x =\frac{1- t ^2}{1+ t ^2}$
When $x =0, t =0$ and when $x =\frac{\pi}{2}, t =1$
$\therefore I =\int_0^1 \frac{1}{5+4\left(\frac{1- t ^2}{1+ t ^2}\right)} \times \frac{2}{1+ t ^2} dt$
$=2 \int_0^1 \frac{1}{5+5 t +4-4 t ^2} dt$
$=2 \int_0^1 \frac{1}{9+ t ^2} dt$
$=2 \int_0^1 \frac{1}{ t ^2+3^2} dt$
$=2\left[\frac{1}{3} \tan ^{-1}\left(\frac{ t }{3}\right)\right]_0^1$
$=\frac{2}{3}\left[\tan ^{-1}\left(\frac{1}{3}\right)-\tan ^{-1}(0)\right]$
$=\frac{2}{3} \tan ^{-1}\left(\frac{1}{3}\right)$
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