Question

$\int \frac{1}{8{\sin }^{2}x+1}$ dx is equal to

• A. $\sin -1(\tan x)+C$
• B. $\frac{1}{3}{\sin }^{-1}\left(\tan x\right)+C$
• C. $\frac{1}{3}{\tan }^{-1}\left(3\tan x\right)+C$
• D. ${\tan }^{-1}(3tanx)+C$
• E. ${\sin }^{-1}(3\tan x)+C$

Answer 1

Answer: (C)

Le $I=\int \frac{1}{8{\sin }^{2}x+1}$
$=\int \frac{{\sec }^{2}x}{8{\tan }^{2}x+{\sec }^{2}x}dx$
[dividing numerator and denominator by ${\cos }^{2}x]$
$=\int \frac{{\sec }^{2}x}{1+(3\tan x{)}^2}dx$
Let $t=3\tan x$
$\Rightarrow \frac{dt}{dx}=3{\sec }^{2}x$
$\Rightarrow \frac{dt}{3}={\sec }^{2}xdx$
$\therefore I=\frac{1}{3}\int \frac{dt}{1+(t{)}^2}=\frac{1}{3}{\tan }^{-1}(t)+C$
$=\frac{1}{3}{\tan }^{-1}(3\tan x)+C$