Question

${\int }_0^{\frac{\pi }{2}}\frac{1}{4co{s}^{2}x+9si{n}^{2}x}dx=$

• A. $\frac{\pi }{12}$
• B. $\frac{\pi }{10}$
• C. $\frac{\pi }{5}$
• D. $\frac{\pi }{2}$

Answer 1

The correct option is A $\frac{\pi }{12}$

${\int }_0^{\frac{\pi }{2}}\frac{1}{4co{s}^{2}x+9si{n}^{2}x}dx={\int }_0^{\frac{\pi }{2}}\frac{se{c}^2}{4+9ta{n}^{2}x}dx$
$Puttanx=t.Thense{c}^{2}xdx=dt.x=0,\frac{\pi }{2}\Rightarrow t=0,\infty$
$\therefore {\int }_0^{\frac{\pi }{2}}\frac{se{c}^2}{4+9ta{n}^{2}x}dx={\int }_0^{\infty }\frac{1}{4+9t^2}dt=\frac{1}{3}\frac{1}{2}{\left[Ta{n}^{-1}\frac{3t}{2}\right]}_0^{\infty }=\frac{1}{2}\left[Ta{n}^{-1}\right(\infty )-Ta{n}^{-1}(0\left)\right]$
$=\frac{1}{6}[\frac{\pi }{2}-0]=\frac{\pi }{12}$