Question

${\int }_0^{\pi /2}\frac{d\theta }{9{\sin }^2\theta +4{\cos }^2\theta }=K\pi$ , then K =

• A. $\frac{1}{10}$
• B. $\frac{1}{12}$
• C. $\frac{1}{8}$
• D. $\frac{1}{3}$

Answer 1

Answer: (B)

Given integral $=\underset{0}{\overset{\pi /2}{\int }}\frac{se{c}^2\theta d\theta }{4+9ta{n}^2\theta }$
$=\frac{1}{9}\underset{0}{\overset{\pi /2}{\int }}\frac{se{c}^2\theta d\theta }{{\left(\frac{2}{3}\right)}^2+ta{n}^2\theta }$
Put $tan\theta —z,se{c}^2\theta d\theta =dz$
$=\frac{1}{9}\underset{0}{\overset{\infty }{\int }}\frac{dz}{{\left(\frac{2}{3}\right)}^2+z^2}=\frac{1}{9}\frac{1}{\frac{2}{3}}{\left|ta{n}^{-1}\frac{z}{\frac{2}{3}}\right|}_0^{\infty }$
$=\frac{1}{6}\left(ta{n}^{-1}\infty -ta{n}^{-1}0\right)$
$=\frac{1}{6}\cdot \frac{\pi }{2}=K\pi \left(given\right)\therefore K=\frac{1}{12}$