Question

Integrate :
$\int \frac{1}{3\sin x+4\cos x}dx$

Answer 1

$\int \frac{dx}{3\sin x+4\cos x}$

$=\int \frac{dx}{3\left(\frac{2\tan \frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)+4\left(\frac{1-{\tan }^2\frac{x}{2}}{1+{\tan }^2\frac{x}{2}}\right)}$

$=\int \frac{dx}{3\left(\frac{2\tan \frac{x}{2}}{{\sec }^2\frac{x}{2}}\right)+4\left(\frac{1-{\tan }^2\frac{x}{2}}{{\sec }^2\frac{x}{2}}\right)}$

$=-\int \frac{{\sec }^2\frac{x}{2}dx}{4{\tan }^2\frac{x}{2}-6\tan \frac{x}{2}-4}$

Let $t=\tan \frac{x}{2}\Rightarrow dt=\frac{1}{2}{\sec }^2\frac{x}{2}dx$

$=-2\int \frac{dt}{4t^2-6t-4}$

$=-\int \frac{dt}{2t^2-3t-2}$

$=-\int \frac{dt}{2t^2-4t+t-2}$

$=-\int \frac{dt}{2t(t-2)+(t-2)}$

$=-\int \frac{dt}{(t-2)(2t+1)}$

$=-2\int \frac{dt}{(2t-4)(2t+1)}$

$=\frac{-2}{-5}\int \frac{(2t-4)-(2t+1)}{(2t-4)(2t+1)}dt$

$=\frac{2}{5}\int \frac{(2t-4)-(2t+1)}{(2t-4)(2t+1)}dt$

$=\frac{2}{5}\int \frac{(2t-4)dt}{(2t-4)(2t+1)}-\frac{2}{5}\int \frac{(2t+1)dt}{(2t-4)(2t+1)}$

$=\frac{2}{5}\int \frac{dt}{(2t+1)}-\frac{2}{5}\int \frac{dt}{(2t-4)}$

$=\frac{1}{5}\log |2t+1|-\frac{1}{5}\log |2t-4|+c$

$=\frac{1}{5}\log \left|\frac{2t+1}{2t-4}\right|+c$

$=\frac{1}{5}\log \left|\frac{2\tan \frac{x}{2}+1}{2\tan \frac{x}{2}-4}\right|+c$ where $t=\tan \frac{x}{2}$