Question

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

Answer 1

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

$\Rightarrow 2dt={\sec }^2\frac{x}{2}dx$

$\Rightarrow 2dt=(1+t^2)dx$

$\Rightarrow dx=\frac{2dt}{1+t^2}$

Now, $\cot x=\frac{1-t^2}{1+t^2}$ and $\sin x=\frac{2t}{1+t^2}$

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

$=\int \frac{\frac{2dt}{1+t^2}}{4\left(\frac{1-t^2}{1+t^2}\right)+3\left(\frac{2t}{1+t^2}\right)}$

$=\int \frac{2dt}{4(1-t^2)+6t}$

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

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

$=\int \frac{-dt}{2(t^2-\frac{3}{2}t-1)}$

$=\frac{-1}{2}\int \frac{dt}{t^2-\frac{3}{2}t-1}$

$=\frac{-1}{2}\int \frac{dt}{(t^2-2\times t\times \frac{3}{4}+{\left(\frac{3}{4}\right)}^2)-{\left(\frac{3}{4}\right)}^2-1}$

$=\frac{-1}{2}\int \frac{dt}{{(t-\frac{3}{4})}^2-\frac{25}{16}}$

$=\frac{-1}{2}\times \frac{1}{2\times \frac{5}{4}}\ln \left(\frac{t-\frac{3}{4}}{\frac{5}{4}}\right)$

$=\frac{-1}{5}\ln \frac{4t-3}{5}$ on simplification

$=\frac{-1}{5}\ln \left(\frac{4\tan \frac{x}{2}-3}{5}\right)+c$ where $t=\tan \frac{x}{2}$ and $c$ is the constant of integration.