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. $\tan^{-1} (3\,tan \,x) + C$
• E. $\sin^{-1} ( 3 \tan x) + C$

Answer 1

Answer: (C)

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