Question

If $\int \frac{dx}{5{\cos }^{2}x+4\sin x(\sin x+\cos x)}=\frac{1}{\alpha }{\tan }^{-1}(\frac{1}{\beta }+\gamma \tan x)+C,$ where $C$ is a constant of integration, then $\alpha \beta +\gamma$ is equal to

Answer 1

$I=\int \frac{dx}{5{\cos }^{2}x+4\sin x(\sin x+\cos x)}$
$=\int \frac{dx}{5{\cos }^{2}x+4{\sin }^{2}x+4\sin x\cos x}$
$=\int \frac{{\sec }^{2}xdx}{4{\tan }^{2}x+4\tan x+5}$
Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$I=\int \frac{dt}{4t^2+4t+5}$
$=\frac{1}{4}\int \frac{dt}{t^2+t+\frac{5}{4}}$
$=\frac{1}{4}\int \frac{dt}{{(t+\frac{1}{2})}^2+1}$
$=\frac{1}{4}{\tan }^{-1}(t+\frac{1}{2})+C$
$=\frac{1}{4}{\tan }^{-1}(\frac{1}{2}+\tan x)+C$

$\Rightarrow \alpha =4,\beta =2$ and $\gamma =1$
$\therefore \alpha \beta +\gamma =9$