Evaluate the following integrals:
$\int \frac{1}{5+4 \cos x} d x$
Given $I=\int \frac{1}{5+4 \cos x} d x$
We know that $\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$
$\Rightarrow \int \frac{1}{5+4 \cos x} d x=\int \frac{1}{5+4\left(\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}\right)} d x$
$=\int \frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)+4\left(1-\tan ^{2} \frac{x}{2}\right)} d x$
Replacing $1+\tan ^{2} x / 2$ in numerator by $\sec ^{2} x / 2$,
$\Rightarrow \int \frac{1+\tan ^{2} \frac{x}{2}}{5\left(1+\tan ^{2} \frac{x}{2}\right)+4\left(1-\tan ^{2} \frac{x}{2}\right)} d x=\int \frac{\sec ^{2} \frac{x}{2}}{\tan ^{2} \frac{x}{2}+9} d x$
Putting $\tan x / 2=t$ and $\sec ^{2}(x / 2) d x=2 d t$,
$\Rightarrow \int \frac{\sec ^{2} \frac{x}{2}}{\tan ^{2} \frac{x}{2}+9} d x=\int \frac{2 d t}{t^{2}+9}$
$=2 \int \frac{1}{t^{2}+9} d t$
We know that $\int \frac{1}{\mathrm{a}^{2}+\mathrm{x}^{2}} \mathrm{dx}=\frac{1}{\mathrm{a}} \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)+\mathrm{c}$
$\Rightarrow 2 \int \frac{1}{t^{2}+9} d t=2\left(\frac{1}{3}\right) \tan ^{-1}\left(\frac{t}{3}\right)+c$
$=\frac{2}{3} \tan ^{-1}\left(\frac{\tan x}{3}\right)+c$
$\therefore \mathrm{I}=\int \frac{1}{5+4 \cos \mathrm{x}} \mathrm{dx}=\frac{2}{3} \tan ^{-1}\left(\frac{\tan \mathrm{x}}{3}\right)+\mathrm{c}$
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