Question

Evaluate:-
$\int \frac{\left(3\sin x-2\right)\cos x}{5-{\cos }^{2}x-4\sin x}$

Answer 1

$\int \frac{\left(3\sin x-2\right)\cos x}{5-{\cos }^{2}x-4\sin x}=\int \frac{\left(3\sin x-2\right)\cos x}{4+1-{\cos }^{2}x-4\sin x}=\int \frac{\left(3\sin x-2\right)\cos x}{4+{\sin }^{2}x-4\sin x}=\int \frac{\left(3\sin x-2\right)\cos x}{(\sin x-2{)}^2}$

Now, Let $t=\sin xdt=\cos x$

So,

$\int \frac{\left(3\sin x-2\right)\cos x}{(\sin x-2{)}^2}=\int \frac{3t-2}{(t-2{)}^2}dt$

Using partial fraction we can write

$\frac{3t-2}{(t-2{)}^2}=\frac{A}{(t-2{)}^2}+\frac{B}{t-2}\Rightarrow 3t-2=A+B(t-2)\Rightarrow A=4,B=3$

Hence, $\int \frac{3t-2}{(t-2{)}^2}dt=\int \frac{4}{(t-2{)}^2}+\int \frac{3}{t-2}=-4(t-2{)}^{-1}+3\log (t-2)+c=-4(\sin x-2{)}^{-1}+3\log (\sin x-2)+c$