Evaluate the following intregals: 1 5-4 dx

Question

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

✓

Answer

Let $\text{I}=\int\frac{1}{5-4\cos\text{x}}\ \text{dx}$
Putting $\cos\text{x}=\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$=\text{I}=\int\frac{1}{5-4\Bigg(\frac{1-\tan^2\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\ \text{dx}$
$=\int\frac{\Big(1+\tan^2\frac{\text{x}}{2}\Big)}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)-4+4\tan^2\frac{\text{x}}{2}}$
$=\int\frac{\sec^2\Big(\frac{\text{x}}{2}\Big)}{9\tan^2\frac{\text{x}}{2}+1}\ \text{dx}$
Let $\tan\big(\frac{\text{x}}{2}\big)=\text{t}$
$\Rightarrow\frac{1}{2}\sec^2\big(\frac{\text{x}}{2}\big)\text{dx}=\text{dt}$
$\Rightarrow\sec^2\big(\frac{\text{x}}{2}\big)\text{dx}=2\text{dt}$
$\therefore\text{I}=2\int\frac{\text{dt}}{9\text{t}^2+1}$
$=\frac{2}{9}\int\frac{\text{dt}}{\text{t}^2+\frac{1}{9}}$
$=\frac{2}{9}\int\frac{\text{dt}}{\text{t}^2+\big(\frac{1}{3}\big)^2}$
$=\frac{2}{9}\times3\tan^{-1}\bigg(\frac{\text{t}}{\frac{1}{3}}\bigg)+\text{C}$
$=\frac{2}{3}\tan^{-1}(3\text{t})+\text{C}$
$=\frac{2}{3}\tan^{-1}\big(3\tan\frac{\text{x}}{2}\big)+\text{C}$

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