Question

$\displaystyle\int_{0}^{\pi/2}\frac{d\theta}{9\,\sin^2\theta+4\,\cos^2\theta}=K\,\pi$ , then K =

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

Answer 1

Answer: (B)

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