# Integrate the rational functions$\frac{\cos x}{(1-\sin x)(2-\sin x)}\qquad[Hint:Put\;\sin x=t]$

$\begin{array}{1 1} \log\large \frac{|1-sin x|}{|2-sin x|}+c \\\log\large \frac{|2-sin x|}{|1-sin x|}+c \\\log\large \frac{|1+sin x|}{|2-sin x|}+c \\ \log\large \frac{|1-sin x|}{|2+sin x|}+c \end{array}$

Toolbox:
• $(i)\;$Form of the ration function$\frac{px+q}{(x-a)(x-b)}$
• $\;$Form of the partial function$\frac{A}{(x-a)}+\frac{B}{(x-b)}$
• $(ii)\;\int\frac{dx}{(x-a)}=log|x-a|+c.$
Given:$I=\int\frac{\cos x}{(1-\sin x)(2-\sin x)}dx.$

Put sinx=t.

Therefore $\cos xdx=dt.$

$\int\frac{\cos x}{(1-\sin x)(2-\sin x)}=\int\frac{dt}{(1-t)(2-t)}$

Resolving into partial fraction

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

$\Rightarrow 1=A(2-t)+B(1-t).$

Equating the coefficient of t we get,

0=-A-B-----(1)

1=2A+B-----(2)

Equating the constant term,

1=-A-B------(3)

2A+B=0
-A-B=1
_____________
A=1

Therefore 2A+B=0$\Rightarrow B=-2.$

$I=\int\frac{-2}{(1-t)}+\int\frac{dt}{2-t}.$

$\;\;\;=-2\int\frac{dt}{(1-t)}+\int\frac{dt}{(2-t)}.$

On integrating we get,

$\;\;\;=-2log|1-t|+log|2-t|.$

substituting for t we get,

$\;\;\;=-2log|1-\sin x|+log|2-\sin x|.$

edited Feb 7, 2013