Question

${\int }_0^{\pi }\frac{dx}{3+2\sin x+\cos x}=$

• A. $\pi /3$
• B. $\pi /4$
• C. $\pi /6$
• D. $\pi /2$

Answer 1

Answer: (B)

$\pi /4$

${\int }_0^{\pi }\frac{dx}{3+2sinx+cosx}$
$={\int }_0^{\pi }\frac{dx(1+ta{n}^2\frac{x}{2})}{3(1+ta{n}^2\frac{x}{2})+2\left(2tan\frac{x}{2}\right)+(1-ta{n}^2\frac{x}{2})}$
$={\int }_0^{\pi }\frac{se{c}^2\frac{x}{2}dx}{2ta{n}^2\frac{x}{2}+4tan\frac{x}{2}+4}$
$={\int }_0^{\pi }\frac{2d\left(ta{n}^2\frac{x}{2}\right)}{(\sqrt{2}tan\frac{x}{2}+\sqrt{2}{)}^2+(\sqrt{2}{)}^2}$
$={\int }_0^{\pi }\frac{d\left(ta{n}^2\frac{x}{2}\right)}{(\sqrt{2}tan\frac{x}{2}+\sqrt{2}{)}^2+(\sqrt{2}{)}^2}$
$={\int }_0^{\pi }\frac{d\left(ta{n}^2\frac{x}{2}\right)}{(tan\frac{x}{2}+1{)}^2}=tn{a}^{-1}(tan\frac{x}{2}+1){\int }_0^{\pi }$
$=\frac{\pi }{2}-\frac{\pi }{4}$
$=\frac{\pi }{4}$