Evaluate the integral:


Evaluate the integral:

$\int \frac{(3 \sin x-2) \cos x}{5-\cos ^{2} x-4 \sin x} d x$


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

$\Rightarrow I=\int \frac{(3 \sin x-2) \cos x}{4+\sin ^{2} x-4 \sin x} d x$

Let, $\sin x=t \Rightarrow \cos x d x=d t$

$\therefore I=\int \frac{(3 t-2)}{t^{2}-4 t+4} d t$

As we can see that there is a term of $t$ in numerator and derivative of $t^{2}$ is also $2 t$. So there is a chance that we can make substitution for $t^{2}-4 t+4$ and $I$ can be reduced to a fundamental integration.

As, $\frac{d}{d t}\left(t^{2}-4 t-4\right)=2 t-4$

$\therefore$ Let, $3 t-2=A(2 t-4)+B$

$\Rightarrow 3 t-2=2 A t-4 A+B$

On comparing both sides -

We have,

$2 A=3 \Rightarrow A=3 / 2$

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


$I=\int \frac{(3 t-2)}{t^{2}-4 t+4} d t$

$\therefore I=\int \frac{\frac{3}{2}(2 t-4)}{t^{2}-4 t+4} d t+\int \frac{4}{t^{2}-4 t+4} d t$

Let, $I_{1}=\frac{3}{2} \int \frac{(2 t-4)}{t^{2}-4 t+4} d t$ and $I_{2}=\int \frac{4}{t^{2}-4 t+4} d t$

Now, $I=I_{1}+I_{2} \ldots$ eqn 1

We will solve $I_{1}$ and $I_{2}$ individually.

$\mathrm{As}, \mathrm{I}_{1}=\frac{3}{2} \int \frac{(2 \mathrm{t}-4)}{\mathrm{t}^{2}-4 \mathrm{t}+4} \mathrm{dt}$

Let $u=t^{2}-4 t+4 \Rightarrow d u=(2 t-4) d x$

$\therefore I_{1}$ reduces to $\frac{3}{2} \int \frac{d u}{u}$


$\mathrm{I}_{1}=\frac{3}{2} \int \frac{\mathrm{du}}{\mathrm{u}}=\log |\mathrm{u}|+\mathrm{C}\left\{\because \int \frac{\mathrm{dx}}{\mathrm{x}}=\log |\mathrm{x}|+\mathrm{C}\right\}$

On substituting value of $u$, we have:

$\mathrm{I}_{1}=\frac{3}{2} \log \left|\mathrm{t}^{2}-4 \mathrm{t}+4\right|+\mathrm{C}$

$I_{1}=\frac{3}{2} \log |t-2|^{2}+C=3 \log |t-2|+C \ldots .$ eqn 2

$\because I_{2}=\int \frac{4}{t^{2}-4 t+4} d t$

$\Rightarrow I_{2}=\int \frac{4}{\left\{t^{2}-2(2) t+2^{2}\right\}} d x$

Using: $a^{2}-2 a b+b^{2}=(a-b)^{2}$

We have:

$I_{2}=4 \int \frac{1}{(t-2)^{2}} d x$

As, $\int \frac{1}{x^{2}} d x=-\frac{1}{x}$

$\therefore I_{2}=\frac{-4}{t-2}=\frac{4}{2-t}+C \ldots$ eqn 3

From eqn 1, we have:


Using eqn 2 and 3 , we get -

$I=3 \log |t-2|+\frac{4}{2-t}+C$

Putting value of $t$ in $l$ :

$I=3 \log |\sin x-2|+\frac{4}{2-\sin x}+C \ldots \ldots$ ans

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