Question

$\int_{}^{}\frac{dx}{4\sin^{2}x + 4\sin x.\cos x + 5\cos^{2}x}$ equals

• A. $\tan^{- 1}\left( \tan x + \frac{1}{2} \right) + c$
• B. $\frac{1}{4}\tan^{- 1}\left( \tan x + \frac{1}{2} \right) + c$
• C. $4\tan^{- 1}\left( \tan x + \frac{1}{2} \right) + c$
• D. None of these

Answer 1

Answer: (B)

$\frac{1}{4}\tan^{- 1}\left( \tan x + \frac{1}{2} \right) + c$

Answer 2

I=$\int_{}^{}\frac{dx}{4\sin^{2}x + 4\sin x.\cos x + 5\cos^{2}x}$ $= \int_{}^{}\frac{\sec^{2}xdx}{4\tan^{2}x + 5 + 4\tan x}$ $= \frac{1}{4}\int_{}^{}\frac{\sec^{2}xdx}{\left( \tan x + \frac{1}{2} \right)^{2} + 1}$

Put $\tan x + \frac{1}{2} = t$ $\Rightarrow \sec^{2}xdx = dt = \frac{1}{4}\int_{}^{}\frac{dt}{t^{2} + 1} = \frac{1}{4}\tan^{- 1}t + c$ ⇒ I $= \frac{1}{4}\int_{}^{}{\tan^{- 1}\left( \tan x + \frac{1}{2} \right)} + c$.