Question

${\int }_0^{\pi }\frac{1}{3+2sinx+cosx}dx=$

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is B $\frac{\pi }{4}$

$Putt=\frac{tanx}{2}.Thendt=se{c}^2\frac{x}{2}.\frac{1}{2}dx\Rightarrow dx=\frac{2dt}{1+t^2};x=0,\pi \Rightarrow t=0,\infty sinx=\frac{2t}{1+t^2},cosx=\frac{1-t^2}{1+t^2}$
$\therefore {\int }_0^{\pi }\frac{1}{3+2sinx+cosx}dx={\int }_0^{\infty }\frac{1}{3+2\left[\frac{2t}{(1+t^2)+\left[\frac{(1-t^2)}{(1+t^2)}\right]}\right]}\frac{2dt}{1+t^2}$
$={\int }_0^{\infty }\frac{1+t^2}{3+3t^2+4t+1-t^2}\frac{2dt}{1+t^2}=2{\int }_0^{\infty }\frac{1}{2t^2+4t+4}dt={\int }_0^{\infty }\frac{1}{t^2+2t+2}dt$
$={\int }_0^{\infty }\frac{1}{(t+1{)}^2+1^2}dt=\left[Ta{n}^{-1}\right(t+1){]}_0^{\infty }=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$