Let I = 1 2+ x- x d x Put (x 2 )= t x =2 ^ -1 t dx =2 dt 1+ t ^2 and x =2 t 1+ t ^2 , x =1- t ^2 1+ t ^2 I= 1 2+ (1- t ^2 1+ t ^2 )-2 t 1+ t ^2 2 dt 1+ t ^2 = 2 2+2 t ^2+1- t ^ -2 t dt =2 1 t ^2-2 t +3 dt (1 2 coefficient of t )^2= (1 2 -2 )^2 =(-1)^2 =1 I =2 1 t ^2-2 t +1-1+3 dt =2 1 ( t -1)^2+2 dt =2 1 ( t -1)^2+(2)^2 dt =2 1 2 ^ -1 ( t -1 2 )+ c I =2 ^ -1 [ (x 2 )-1 2 ]+ c

1 2+ x- x d x

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Question

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

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Answer

$\text { Let } I =\int \frac{1}{2+\cos x-\sin x} d x$
$\text { Put } \tan \left(\frac{x}{2}\right)= t$
$\therefore x =2 \tan ^{-1} t$
$\therefore dx =\frac{2 dt }{1+ t ^2} \text { and } \sin x =\frac{2 t }{1+ t ^2}, \cos x =\frac{1- t ^2}{1+ t ^2}$
$\therefore I=\int \frac{1}{2+\left(\frac{1- t ^2}{1+ t ^2}\right)-\frac{2 t }{1+ t ^2}} \times \frac{2 dt }{1+ t ^2}$
$=\int \frac{2}{2+2 t ^2+1- t ^{-2} t } dt$
$=2 \int \frac{1}{ t ^2-2 t +3} dt$
$\left(\frac{1}{2} \text { coefficient of } t \right)^2=\left(\frac{1}{2} \times-2\right)^2$
$=(-1)^2$
$=1$
$\therefore I =2 \int \frac{1}{ t ^2-2 t +1-1+3} dt$
$=2 \int \frac{1}{( t -1)^2+2} dt$
$=2 \int \frac{1}{( t -1)^2+(\sqrt{2})^2} dt$
$=2 \cdot \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{ t -1}{\sqrt{2}}\right)+ c$
$\therefore I =\sqrt{2} \tan ^{-1}\left[\frac{\tan \left(\frac{x}{2}\right)-1}{\sqrt{2}}\right]+ c $