Evaluate the following intregals: 1 5-4 dx - Vidyadip

Question

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

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Answer

Let $\text{I}=\int\frac{1}{5-4\sin\text{x}}\ \text{dx}$
Put $\sin\text{x}=\frac{2\tan^\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}$
$\text{I}=\int\frac{1}{5-4\Bigg(\frac{2\tan\frac{\text{x}}{2}}{1+\tan^2\frac{\text{x}}{2}}\Bigg)}\ \text{dx}$
$=\int\frac{1+\tan^2\frac{\text{x}}{2}}{5\Big(1+\tan^2\frac{\text{x}}{2}\Big)-4\Big(2-\tan\frac{\text{x}}{2}\Big)}\ \text{dx}$
$=\int\frac{\sec^2\frac{\text{x}}{2}}{5+5\tan^2\frac{\text{x}}{2}-8\tan\frac{\text{x}}{2}}\ \text{dx}$
Let $\tan\frac{\text{x}}{2}=\text{t}$
$\frac{1}{2}\sec^2\frac{\text{x}}{2}\text{dx}=\text{dt}$
$=\int\frac{2\text{dt}}{5\text{t}^2-8\text{t}+5}$
$=\frac{2}{5}\int\frac{\text{dt}}{\text{t}^2-2\text{t}\Big(\frac{4}{5}\Big)+\Big(\frac{4}{5}\Big)^2-\Big(\frac{4}{5}\Big)^2+1}$
$\text{I}=\frac{2}{5}\int\frac{\text{dt}}{\Big(\text{t}-\frac{4}{5}\Big)^2+\Big(\frac{3}{5}\Big)^2}$
$=\frac{2}{5}\times\frac{1}{\frac{3}{5}}\Bigg(\frac{\text{t}-\frac{4}{5}}{\frac{3}{5}}\Bigg)+\text{C}$
$=\frac{2}{3}\tan^{-1}\Big(\frac{5\text{t}-4}{3}\Big)+\text{C}$
$\text{I}=\frac{2}{3}\tan^{-1}\Big(\frac{5\tan\frac{\text{x}}{2}-4}{3}\Big)+\text{C}$

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