Question

Solve: $\int \frac{1}{(\sin x-2\cos x)(2\sin x+\cos x)}dx$

Answer 1

Let, $I=\int \frac{1}{(\sin x-2\cos x)(2\sin x+\cos x)}dx=\int \frac{1}{2{\sin }^{2}x-3\sin x\cos x-2{\cos }^{2}x}dx=\int \frac{1}{-2({\cos }^{2}x-{\sin }^{2}x)-3\sin x\cos x}dx=-\int \frac{1}{2\cos 2x+\frac{3}{2}\sin 2x}dx$

Now, let $\tan x=t\therefore dx=\frac{2}{1+t^2}dt,\cos 2x=\frac{1-t^2}{1+t^2},\sin 2x=\frac{2t}{1+t^2}$

We can write,

$I=-\int \frac{1}{2\left(\frac{1-t^2}{1+t^2}\right)+\frac{3}{2}\left(\frac{2t}{1+t^2}\right)}\cdot \left(\frac{2}{1+t^2}dt\right)=-\int \frac{1}{\frac{2-2t^2+3t}{1+t^2}}\cdot \frac{2}{1+t^2}dt=\int \frac{2}{2t^2-3t-2}dt=\int \frac{1}{t^2-\frac{3}{2}t-1}dt=\int \frac{1}{t^2-\frac{3}{2}t-1+\frac{9}{16}-\frac{9}{16}}dt=\int \frac{1}{(t^2-\frac{3}{2}t+\frac{9}{16})-(\frac{9}{16}+\frac{16}{16})}dt[putting,1=\frac{16}{16}]=\int \frac{1}{(t-\frac{3}{2}{)}^2-{\left(\frac{5}{4}\right)}^2}dt$

We know that,

$\int \frac{1}{(x^2-a^2)}dx=\frac{1}{2a}\log \left[\frac{x-a}{x+a}\right]+C$

$\therefore I=\frac{1}{2\left(\frac{5}{4}\right)}\log \left[\frac{(t-\frac{3}{2})-\frac{5}{4}}{(t-\frac{3}{2})+\frac{5}{4}}\right]+C=\frac{2}{5}\log \left[\frac{4t-11}{4t-1}\right]+C$

Putting $t=\tan x,\Rightarrow I=\frac{2}{5}\log \left[\frac{4\tan x-11}{4\tan x-1}\right]+C\therefore \int \frac{1}{(\sin x-2\cos x)(2\sin x+\cos x)}dx=\frac{2}{5}\log \left[\frac{4\tan x-11}{4\tan x-1}\right]+C.$