Evaluate : d x 2+ x- x - Vidyadip

Question

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

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Answer

$I =\int \frac{d x}{2+\cos x-\sin x}$
Put $\tan \left(\frac{x}{2}\right)=t, \therefore \frac{x}{2}=\tan ^{-1} t$
$\therefore x=2 \tan ^{-1} t, d x=\frac{2}{1+t^2} d t$
Also $\cos x=\frac{1-t^2}{1+t^2}, \sin x=\frac{2 t}{1+t^2}$
$\therefore I =\int \frac{\frac{2 d t}{1+t^2}}{2+\frac{1-t^2}{1+t^2}-\frac{2 t}{1+t^2}}$
$ =\int \frac{2 d t}{2+2 t^2+1-t^2-2 t}$
$ =\int \frac{2 d t}{t^2-2 t+3}$
$ =2 \int \frac{d t}{t^2-2 t+1+2}$
$ =2 \int \frac{d t}{(t-1)^2+(\sqrt{2})^2}$
$=\frac{2}{\sqrt{2}} \tan ^{-1}\left(\frac{t-1}{\sqrt{2}}\right)+c$
$=\sqrt{2} \tan ^{-1}\left(\frac{\tan \frac{x}{2}-1}{\sqrt{2}}\right)+c$

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