Question

${\int }_0^{\pi /2}\frac{dx}{4{\cos }^{2}x+9{\sin }^{2}x}=$

• A. $\pi /6$
• B. $\pi /12$
• C. $\pi /36$
• D. $\pi /72$

Answer 1

The correct option is B

$\pi /12$

${\int }_0^{\pi /2}\frac{dx}{4{\cos }^{2}x+9{\sin }^{2}x}$

$={\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{4+9{\tan }^{2}x}$

Put $3\tan x=t$
$\therefore {\int }_0^{\pi /2}\frac{{\sec }^{2}xdx}{4+9{\tan }^{2}x}=\frac{1}{3}{\int }_0^{\infty }\frac{dt}{4+t^2}=\frac{\pi }{12}$