Question

If $\int \frac{1}{1+{\cos }^{2}x+2\cos x\sin x}dx={\tan }^{-1}\left[f\right(x\left)\right]+C,$ then $f(\frac{\pi }{4}-x)$ is equal to
(where $C$ is integration constant)

• A. $\frac{1}{1+\tan x}$
• B. $\frac{1}{1+\cot x}$
• C. $\frac{2}{1+\cot x}$
• D. $\frac{2}{1+\tan x}$

Answer 1

The correct option is D $\frac{2}{1+\tan x}$

$I=\int \frac{1}{1+{\cos }^{2}x+2\cos x\sin x}dx$
Dividing numerator and denominator by ${\cos }^{2}x$
$\Rightarrow I=\int \frac{{\sec }^{2}x}{{\sec }^{2}x+1+2\tan x}dx$
Put $\tan x=t$
$\Rightarrow {\sec }^{2}xdx=dt\Rightarrow I=\int \frac{dt}{t^2+2t+2}=\int \frac{dt}{(1{)}^2+(t+1{)}^2}\Rightarrow I={\tan }^{-1}(t+1)+C\Rightarrow I={\tan }^{-1}(\tan x+1)+C\Rightarrow f\left(x\right)=1+\tan x\Rightarrow f(\frac{\pi }{4}-x)=1+\tan (\frac{\pi }{4}-x)=1+\frac{1-\tan x}{1+\tan x}\Rightarrow f(\frac{\pi }{4}-x)=\frac{2}{1+\tan x}$