explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 60

$\frac{1}{10} \log (\cos x-1)-\frac{1}{2} \log (\cos +1)+\frac{2}{5} \log \left(\frac{3}{2}+\cos x\right)+C$

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

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

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

$I=\frac{1}{2} \int \frac{1}{(t-1)(t+1)\left(\frac{3}{2}+t\right)} d t$                (1)

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

\begin{aligned} &\frac{1}{(t-1)(t+1)\left(\frac{3}{2}+t\right)}=\frac{A(t+1)\left(\frac{3}{2}+t\right)+B(t-1)\left(\frac{3}{2}+t\right)+C\left(t^{2}-1\right)}{(t-1)(t+1)\left(\frac{3}{2}+t\right)} \\ &1=A(t+1)\left(\frac{3}{2}+t\right)+B(t-1)\left(\frac{3}{2}+t\right)+C\left(t^{2}-1\right) \end{aligned}

\begin{aligned} &\text { At } t=-1 \\ &1=-2 B\left(\frac{1}{2}\right) \\ &B=-1 \\ &\text { At } t=1 \\ &1=2 A\left(\frac{5}{2}\right) \end{aligned}

\begin{aligned} &A=\frac{1}{5} \\ &\text { At } t=\frac{-3}{2} \\ &1=C\left(\frac{9}{4}-1\right) \\ &C=\frac{4}{5} \end{aligned}

Put in (1)

\begin{aligned} &I=\frac{1}{2} \times \frac{1}{5} \int \frac{1}{t-1} d t+\frac{(-1)}{2} \int \frac{1}{t+1} d t+\frac{1}{2} \times \frac{4}{5} \int \frac{1}{\frac{3}{2}+t} d t \\ &I=\frac{1}{10} \log (t-1)-\frac{1}{2} \log (t+1)+\frac{2}{5} \log \left(\frac{3}{2}+t\right)+C \\ &I=\log (\cos x-1)-\frac{1}{2} \log (\cos x+1)+\frac{2}{5} \log \left(\frac{3}{2}+\cos x\right)+C \end{aligned}