Question

$\int_{}^{}\frac{dx}{1 + 3\sin^{2}x} =$

• A. $\frac{1}{3}\tan^{- 1}(3\tan^{2}x) + c$
• B. $\frac{1}{2}\tan^{- 1}(2\tan x) + c$
• C. $\tan^{- 1}(\tan x) + c$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{2}\tan^{- 1}(2\tan x) + c$

Answer 2

$\int_{}^{}\frac{\sec^{2}xdx}{\sec^{2}x + 3\tan^{2}x} = \int_{}^{}\frac{\sec^{2}xdx}{1 + 4\tan^{2}x}$Put $2\tan x = t$

$\sec^{2}xdx = \frac{dt}{2}$

$\therefore I = \frac{1}{2}\int_{}^{}{\frac{dt}{1 + t^{2}} = \frac{1}{2}\tan^{- 1}t + c}$

$\Rightarrow$ $I = \frac{1}{2}\tan^{- 1}(2\tan x) + c$.