# What is the integration? 1/(2sin x+ cos x +1)

Solution:  We have ,

$I=\int \frac{1}{2\sin x+\cos x+1}dx$

Put ,     $\tan (\frac{x}{2})=t\Rightarrow dx=\frac{2}{1+t^2}dt$

$\\ \Rightarrow \cos x=(\frac{1-t^{2}}{1+t^{2}});\sin x=(\frac{2t}{1+t^{2}})\\ \\\Rightarrow 2\sin x+\cos x+1=2(\frac{2t}{1+t^{2}})+(\frac{1-t^{2}}{1+t^{2}})+1=\frac{2(1+2t)}{1+t^{2}}$

$\therefore \int\frac{1}{(1+2t)}dt=\frac{1}{2}\ln(1+2t)+c$

$\Rightarrow I=\frac{1}{2}\ln[1+2\tan(\frac{x}{2})]+c.$

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