Question

Integrate
$\int \frac{2\sin 2x-\cos x}{6-{\cos }^{2}x-4\mathrm{sinx}}dx$

Answer 1

$I=\int \frac{(2\sin 2x-\cos x)}{6-{\cos }^{2}x-4\sin x}dx$

$=\int \frac{(4\sin x\cos x-\cos x)}{6-{\cos }^{2}x-4\sin x}dx$

$=\int \frac{\cos x(4\sin x-1)}{6-1+{\sin }^{2}x-4\sin x}dx$

$=\int \frac{\cos x(4\sin x-1)}{{\sin }^{2}x-4\sin x+5}dx$

Put $t=\sin x\Rightarrow dt=\cos xdx$

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

Let $4t-1=A\times \frac{d}{dt}(t^2-4t+5)+B$

$4t-1=A(2t-4)+B$

$\Rightarrow 2A=4,-4A+B=-1$

$\Rightarrow A=2,B=-1+4A=-1+4\times 2=-1+8=7$

$\therefore A=2,B=7$

So,$4t-1=2(2t-4)+7$

So,$I=\int \frac{[2(2t-4)+7]dt}{t^2-4t+5}$

$=2\int \frac{(2t-4)dt}{t^2-4t+5}+7\int \frac{dt}{t^2-4t+5}$

$I=I_1+7\int \frac{dt}{t^2-4t+5}$

Let $I_1=2\int \frac{(2t-4)dt}{t^2-4t+5}$

Take $v=t^2-4t+5\Rightarrow dv=(2t-4)dt$

$I_1=2\int \frac{dv}{v}=2\log \left|v\right|$

$=\log (t^2-4t+5)$ where $v=t^2-4t+5$

$=\log ({\sin }^{2}x-4\sin x+5)$ where $t=\sin x$

$I=\log ({\sin }^{2}x-4\sin x+5)+7\int \frac{dt}{t^2-4t+4+1}$

$I=\log ({\sin }^{2}x-4\sin x+5)+7\int \frac{dt}{{(t-2)}^2+1}$

$I=\log ({\sin }^{2}x-4\sin x+5)+7{\tan }^{-1}(t-2)+c$

$I=\log ({\sin }^{2}x-4\sin x+5)+7{\tan }^{-1}(\sin x-2)+c$

where $t=\sin x$