Evaluate the following integrals:
Question:

Evaluate the following integrals:

$\int \frac{1}{5+4 \cos x} d x$

Solution:

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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