Question

$\int \frac{dx}{9+16{\sin }^{2}x}$ is equal to

• A. $\frac{1}{3}{\tan }^{-1}\left(\frac{3\tan x}{5}\right)+c$
• B. $\frac{1}{5}{\tan }^{-1}\left(\frac{\tan x}{15}\right)+c$
• C. $\frac{1}{15}{\tan }^{-1}\left(\frac{\tan x}{5}\right)+c$
• D. $\frac{1}{15}{\tan }^{-1}\left(\frac{5\tan x}{3}\right)+c$

Answer 1

The correct option is D $\frac{1}{15}{\tan }^{-1}\left(\frac{5\tan x}{3}\right)+c$

$\int \frac{dx}{9+16{\sin }^{2}x}$
$=\int \frac{dx}{9{\sin }^{2}x+9{\cos }^{2}x+16{\sin }^{2}x}$
$=\int \frac{dx}{9{\cos }^{2}x+25{\sin }^{2}x}$
$=\int \frac{{\sec }^{2}xdx}{9+25{\tan }^{2}x}$
Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$
$=\int \frac{dt}{9+25t^2}=\frac{1}{25}\int \frac{dt}{{\left(\frac{3}{5}\right)}^2+t^2}$
$=\frac{1}{25}\frac{1}{3/5}{\tan }^{-1}\left(\frac{t}{3/5}\right)+c=\frac{1}{15}{\tan }^{-1}\left(\frac{5\tan x}{3}\right)+c$