Evaluate the integral:

$I=\int \frac{(3 \sin x-2) \cos x}{13-\cos ^{2} x-7 \sin x} d x=\int \frac{(3 \sin x-2) \cos x}{13-\left(1-\sin ^{2} x\right)-7 \sin x} d x$

$\Rightarrow I=\int \frac{(3 \sin x-2) \cos x}{12+\sin ^{2} x-7 \sin x} d x$

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

$\therefore I=\int \frac{(3 t-2)}{t^{2}-7 t+12} d t$

As we can see that there is a term of $\mathrm{t}$ in numerator and derivative of $\mathrm{t}^{2}$ is also $2 \mathrm{t}$. So there is a chance that we can make substitution for $t^{2}-7 t+12$ and I can be reduced to a fundamental integration.

As, $\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{t}^{2}-7 \mathrm{t}+12\right)=2 \mathrm{t}-7$

$\therefore$ Let, $3 t-2=A(2 t-7)+B$

$\Rightarrow 3 t-2=2 A t-7 A+B$

On comparing both sides -

We have,

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

$-7 A+B=-2 \Rightarrow B=7 A-2=17 / 2$

Hence,

$I=\int \frac{(3 t-2)}{t^{2}-7 t+12} d t$

$\therefore I=\int \frac{\frac{3}{2}(2 t-7)}{t^{2}-7 t+12} d t+\int \frac{\frac{17}{2}}{t^{2}-7 t+12} d t$

Let, $I_{1}=\frac{3}{2} \int \frac{(2 t-7)}{t^{2}-7 t+12} d t$ and $I_{2}=\frac{17}{2} \int \frac{1}{t^{2}-7 t+12} d t$

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

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

$A S, I_{1}=\frac{3}{2} \int \frac{(2 t-7)}{t^{2}-7 t+12} d t$

Let $u=t^{2}-7 t+12 \Rightarrow d u=(2 t-7) d x$

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

Hence,

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

On substituting value of u, we have:

$I_{1}=\frac{3}{2} \log \left|t^{2}-7 t+12\right|+C \ldots .$ eqn 2

As, $I_{2}=\frac{17}{2} \int \frac{1}{t^{2}-7 t+12}$ dtand we don't have any derivative of function present in denominator. $\therefore$ we will use some special integrals to solve the problem.

i) $\int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C$ ii $\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$

$\because I_{2}=\frac{17}{2} \int \frac{1}{t^{2}-7 t+12} d t$

$\Rightarrow I_{2}=\frac{17}{2} \int \frac{4}{\left\{t^{2}-2\left(\frac{7}{2}\right) t+\left(\frac{7}{2}\right)^{2}\right\}+12-\left(\frac{7}{2}\right)^{2}} d x$

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

We have:

$\mathrm{I}_{2}=\frac{17}{2} \int \frac{1}{\left(\mathrm{t}-\frac{7}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} \mathrm{dx}$

$\mathrm{I}_{2}$ matches with the form $\int \frac{1}{\mathrm{x}^{2}-\mathrm{a}^{2}} \mathrm{dx}=\frac{1}{2 \mathrm{a}} \log \left|\frac{\mathrm{x}-\mathrm{a}}{\mathrm{x}+\mathrm{a}}\right|+\mathrm{C}$

$\therefore I_{2}=\frac{17}{2} \frac{1}{2\left(\frac{1}{2}\right)} \log \left|\frac{\left(t-\frac{7}{2}\right)-\frac{1}{2}}{\left(t-\frac{7}{2}\right)+\frac{1}{2}}\right|+C$

$I_{2}=\frac{17}{2} \log \left|\frac{2 t-7-1}{2 t-7+1}\right|+C=\frac{17}{2} \log \left|\frac{2 t-8}{2 t-6}\right|+C$

$I_{2}=\frac{17}{2} \log \left|\frac{t-4}{t-3}\right|+C \ldots$ eqn 3

From eqn 1, we have:

$I=I_{1}+I_{2}$

Using eqn 2 and 3, we get -

$I=\frac{3}{2} \log \left|t^{2}-7 t+12\right|+\frac{17}{2} \log \left|\frac{t-4}{t-3}\right|+C$

Putting value of $\mathrm{t}$ in $\mathrm{I}$ :

$I=\frac{3}{2} \log \left|\sin ^{2} x-7 \sin x+12\right|+\frac{17}{2} \log \left|\frac{4-\sin x}{3-\sin x}\right|+C \ldots \ldots$ ans