Question

Evaluate: $\int \frac{dx}{{\sin }^{2}x+5\sin x\cos x+2}$

Answer 1

Let $I=\int \frac{dx}{{\sin }^{2}x+5\sin x\cos x+2}dx$

On dividing with ${\cos }^{2}x$, we get

$I=\int \frac{{\sec }^{2}xdx}{{\tan }^{2}x+5\tan x+2{\sec }^{2}x}$

$=\int \frac{{\sec }^{2}xdx}{{\tan }^{2}x+5\tan x+2(1+{\tan }^{2}x)}$

$=\int \frac{{\sec }^{2}xdx}{3{\tan }^{2}x+5\tan x+2}$

Let, $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$

$I=\int \frac{dt}{3t^2+5t+2}$

$I=\int \frac{dt}{(3t+2)(t+1)}$

Now, $\frac{1}{(3t+1)(t+1)}=\frac{A}{(3t+2)}+\frac{B}{(t+1)}$ ... (i)

$1=A(t+1)+B(3t+2)$ .... (ii)

Put $t=-1$

$1=0=B(-3+2)$

$1=B(-1)$

$B=-1$

Put $t=(-\frac{2}{3})$ in (ii), we get

$1=A(\frac{-2}{3}+1)+B\left(0\right)$

$1=A\left(\frac{1}{3}\right)$

$A=3$

$\frac{1}{(3t+2)(t+1)}=\frac{3}{(3t+2)}-\frac{1}{(t+1)}$

$\therefore I=\int [\frac{3}{(3t+2)}-\frac{1}{(t+1)}]dt$

$=\frac{3\log |3t+2|}{3}-\log (t+1)$

$=\log |3t+2|-\log |t+1|$

$=\log \left[\frac{|3t+2|}{|t+1|}\right]$

$=\log \left[\frac{|3\tan x+2|}{|\tan x+1|}\right]$