#### Explain Solution R.D.Sharma Class 12 Chapter 18 Indefinite Integrals Exercise 18.19 Question 14 maths Textbook Solution.

Answer: $10 \ln |\sin x-4|-7 \ln |\sin x-3|+c$

Hint : Let $\sin x=t$

Given: $\int \frac{(3 \sin x-2) \cos x}{13-\cos ^{2} x-7 \sin x} d x$

Solution: $\text { Let } \sin x=t$

$\cos x d x=d t$

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

$\int \frac{(3 t-2) d t}{13-1+t^{2}-7 t}=\int \frac{(3 t-2) d t}{t^{2}-7 t+12}$                                                              $\left[\begin{array}{l} t^{2}-7 t+12=(t-4)(t-3) \\ a x^{2}+b x+c, x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \end{array}\right]$

$\int \frac{3 t-2}{(t-4)(t-3)} d t=\frac{A}{t-4}+\frac{B}{t-3}$                                            ..........(1)

\begin{aligned} &3 t-2=\frac{A}{t-4}+\frac{B}{t-3} \\ &3 t-2=A(t-3)+B(t-4) \\ &3 t-2=A t-3 A+B t-4 B \end{aligned}

On comparing,

$3 \mathrm{t}=\mathrm{At}+\mathrm{Bt} \Rightarrow \mathrm{A}+\mathrm{B}=3 \Rightarrow>\mathrm{A}=3-\mathrm{B}$

$-2=-3 A-4 B=23 A+4 B=2=>3(3-B)+4 B=2=>B=-7$

On solving both
$A=3-(-7) \Rightarrow A=10$

In eqn (1)

\begin{aligned} & \int\left[\frac{10}{t-4}+\frac{-7}{t-3}\right] d t \\ =& \int \frac{10}{t-4} d t-\int \frac{7}{t-3} d t \end{aligned}                                                                    $\int \frac{1}{t} d t=\ln |t|+c$

\begin{aligned} &10 \ln |t-4|-7 l n|t-3|+c \\ &=10 \ln |t-4|-7 \ln |t-3|+c \\ &=10 \ln |\sin x-4|-7 \ln |\sin x-3|+c \end{aligned}