# Get Answers to all your Questions

#### explain solution RD Sharma class 12 Chapter 18 Indefinite Integrals exercise 18.30 question 51

$\log \left|\frac{2-\sin x}{1-\sin x}\right|+C$

Hint:

To solve this integration, we use partial fraction method

Given:

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

Explanation:

Let

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

Let
\begin{aligned} &\sin x=y \\ &\cos x d x=d y \\ &I=\int \frac{d y}{(1-y)(2-y)} \\ &\frac{1}{(1-y)(2-y)}=\frac{A}{1-y}+\frac{B}{2-y} \\ &1=A(2-y)+B(1-y) \end{aligned}

Put $y= 2$

\begin{aligned} &1=A(0)+B(-1) \\ &B=-1 \end{aligned}

Put

\begin{aligned} &y=1 \\ &1=A(1)+B(0) \\ &A=1 \end{aligned}

\begin{aligned} &\frac{1}{(2-y)(1-y)}=\frac{-1}{2-y}+\frac{1}{1-y} \\ &I=\int \frac{1}{1-y} d y-\int \frac{1}{2-y} d y \end{aligned}

\begin{aligned} &I=-\log |1-y|+\log |2-y|+C \\ &I=\log \left|\frac{2-y}{1-y}\right|+C \\ &I=\log \left|\frac{2-\sin x}{1-\sin x}\right|+C \end{aligned}                                 $\quad[y=\sin x]$