Question

$\int \frac{1+{\cos }^{2}x}{1+\cos 2x}dx=$

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

Answer 1

Answer: (D)

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

$\int \frac{1+{\cos }^{2}x}{1+\cos 2x}.dx$
$=\int \frac{1+{\cos }^{2}x}{1+{\cos }^{2}x-{\sin }^{2}x}dx$
$=\int \frac{1+{\cos }^{2}x}{(1-{\sin }^{2}x)+{\cos }^{2}x}dx$
$=\int \frac{1+{\cos }^{2}x}{2{\cos }^{2}xdx}$
$=\int \frac{1}{2{\cos }^{2}x}+\frac{{\cos }^{2}x}{2{\cos }^{2}x}dx$
$=\int (\frac{{\sec }^{2}x}{2}+\frac{1}{2})dx$
$=\frac{\tan x+x}{2}+c$