#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 59

$\frac{1}{18} \log (1+\sin x)-\frac{1}{2} \log (1-\sin x)+\frac{4}{9} \log \left(\frac{5}{4}-\sin x\right)+C$

Hint:

To solve this integration, we use partial fraction method

Given:

$\int \frac{1}{\cos x(5-4 \sin x)} d x$

Explanation:

Let

$I=\int \frac{1}{\cos x(5-4 \sin x)} d x$

\begin{aligned} &\text { Put }\\ &\sin x=t\\ &\cos x d x=d t\\ &d x=\frac{1}{\cos x} d t \end{aligned}

\begin{aligned} &I=\int \frac{1}{\cos ^{2} x(5-4 t)}=\int \frac{1}{\left(1-\sin ^{2} x\right)(5-4 t)} d t \\ &I=\int \frac{1}{\left(1-t^{2}\right)(5-4 t)} d t \\ &I=\frac{1}{4} \int \frac{1}{(1+t)(1-t)\left(\frac{5}{4}-t\right)} d t \end{aligned}                        (1)

$\frac{1}{(1+t)(1-t)\left(\frac{5}{4}-t\right)}=\frac{A}{1+t}+\frac{B}{1-t}+\frac{C}{\frac{5}{4}-t}$

\begin{aligned} &\frac{1}{(1+t)(1-t)\left(\frac{5}{4}-t\right)}=\frac{\left[A(1-t)\left(\frac{5}{4}-t\right)+B(1+t)\left(\frac{5}{4}-t\right)+C\left(1-t^{2}\right)\right]}{(1+t)(1-t)\left(\frac{5}{4}-t\right)} \\ &1=A(1-t)\left(\frac{5}{4}-t\right)+B(1+t)\left(\frac{5}{4}-t\right)+C\left(1-t^{2}\right) \end{aligned}

\begin{aligned} &\text { Put } t=1 \\ &1=2 B\left(\frac{1}{4}\right) \\ &B=2 \end{aligned}

$\text { Put } t=-1 \\$

$1=A(2)\left(\frac{9}{4}\right)$

$A=\frac{2}{9}$

\begin{aligned} &\text { Put } t=\frac{5}{4} \\ &1=C\left(1-\frac{25}{16}\right) \\ &1=\frac{-9}{16} C \end{aligned}

\begin{aligned} &C=\frac{-16}{9} \\ &I=\frac{1}{4} \int \frac{\frac{2}{9}}{1+t} d t+\frac{1}{4} \int \frac{2}{(1-t)} d t-\frac{1}{4} \times \frac{16}{9} \int \frac{1}{\frac{5}{4}-t} d t \\ &I=\frac{1}{18} \log (1+t)-\frac{1}{2} \log (1-t)+\frac{4}{9} \log \left(\frac{5}{4}-t\right)+C \\ &I=\frac{1}{18} \log (1+\sin x)-\frac{1}{2} \log (1-\sin x)+\frac{4}{9} \log \left(\frac{5}{4}-\sin x\right)+C \end{aligned}