Question
Evaluate the following intregals:
$\int\frac{1}{5+4\cos\text{x}}\ \text{dx}$

Answer

Let $\text{I}=\int\frac{1}{5+4\cos\text{x}}\ \text{dx}$
Put $\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=\int\frac{1}{5+4\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\ \text{dx}$
$=\int\frac{1+\tan^2\frac{\text{x}}{2}}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)+4\Big(1-\tan^2\frac{\text{x}}{2}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{9+\tan^2\frac{\text{x}}{2}}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$=\int\frac{2\text{dt}}{(3)^2+\text{t}^2}$
$=2\times\frac{1}{3}\tan^{-1}(\text{t})+\text{C}$
$\text{I}=\frac{2}{3}\tan^{-1}\Big(\frac{\tan\frac{\text{x}}{2}}{3}\Big)+\text{C}$

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