Question

$\int \frac{dx}{4{\sin }^{2}x+5{\cos }^{2}x}=?$

• A. $\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{2\tan x}{\sqrt{5}}\right)+c$
• B. $\frac{1}{\sqrt{5}}{\tan }^{-1}\left(\frac{\tan x}{\sqrt{5}}\right)+c$
• C. $\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{2\tan x}{\sqrt{5}}\right)+c$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{2\tan x}{\sqrt{5}}\right)+c$

Let $I=\int \frac{dx}{4\sin {}^{2}x+5\cos {}^{2}x}$

$I=\int \frac{dx\times \frac{1}{4\cos {}^{2}x}}{\tan {}^{2}x+\frac{5}{4}}I=\frac{1}{4}\int \frac{\sec {}^{2}xdx}{\tan {}^{2}x+\frac{5}{4}}$

Let $\tan x=z\sec {}^{2}xdx=dz$

$I=\frac{1}{4}\int \frac{dz}{z^2+\frac{5}{4}}I=\frac{1}{4}\times \frac{1}{\sqrt{\frac{5}{4}}}{\tan }^{-1}\left(\frac{z}{\sqrt{\frac{5}{4}}}\right)+CI=\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{2z}{\sqrt{5}}\right)+C$

and as $z=\tan x$

$I=\frac{1}{2\sqrt{5}}{\tan }^{-1}\left(\frac{2\tan x}{\sqrt{5}}\right)+C$