#### Please solve RD Sharma Class 12 Chapter Indefinite Integral Exercise Rivision Exercise Question 66 maths textbook solution.

$I=+\frac{1}{2} \ln |1+\cos x|-\frac{1}{10} \ln |1-\cos x|-\frac{3}{5} \ln |3 \cos x+2|+c$

Hint: To solve the given solution we multiply and divide the given statement with sin x.

Given :  $\int \frac{1}{\sin x(2+3 \cos x)} d x$

Solution :

\begin{aligned} &I=\int \frac{1}{\sin x(2+3 \cos x)} d x \\ &I=\int \frac{\sin x}{\sin ^{2} x(2+3 \cos x)} d x \\ &I=\int \frac{\sin x d x}{\left(1-\cos ^{2} x\right)(2+3 \cos x)} \end{aligned}

\begin{aligned} &t=\cos x, d t=-\sin x d x \\ &I=\int \frac{-d t}{\left(1-t^{2}\right)(2+3 t)} \\ &I=\int \frac{-d t}{(1+t)(1-t)(2+3 t)} \end{aligned}

$\\\frac{1}{(1+t)(1-t)(2-3 t)}=\frac{A}{1+t}+\frac{B}{1-t}+\frac{c}{2+3 t}=A(1-t)(2+3 t)+B(1+t)(2+3 t)+C(1+t)(1-t)$

$\begin{array}{ll} A=\frac{1}{(1-(-1))(2-3)}=-\frac{1}{2} &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because 1+t=0, t=-1]} \\ B=\frac{1}{(1+1)(2+3)}=\frac{1}{10} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because 1-t=0, t=1]} \\ C=\frac{1}{1-\left(\frac{4}{9}\right)}=\frac{9}{5} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {\left[\because 2+3 t=0, t=-\frac{2}{3}\right]} \end{array}$

\begin{aligned} &I=-\int d t\left(\frac{A}{1+t}+\frac{B}{1-t}+\frac{C}{2+3 t}\right) \\ &I=-\left[A \int \frac{1}{1+t} d t+B \int \frac{1}{1-t} d t+C \int \frac{1}{2+3 t} d t\right. \end{aligned}

\begin{aligned} &I=-\left[-\frac{1}{2} \ln |1+t|+\frac{1}{10} \ln |1-t|+\frac{9}{5} \times \frac{1}{3} \ln |2+3 t|+c\right. \\ &I=+\frac{1}{2} \ln |1+t|-\frac{1}{10} \ln |1-t|-\frac{3}{5} \ln |3 t+2|+c \end{aligned}

$I=+\frac{1}{2} \ln |1+\cos x|-\frac{1}{10} \ln |1-\cos x|-\frac{3}{5} \ln |3 \cos x+2|+c$