Integral Calculus Question 365
Question: $ \int_{{}}^{{}}{e^{2x}\frac{1+\sin 2x}{1+\cos 2x}}\ dx= $
Options:
A) $ e^{2x}\tan x+c $
B) $ e^{2x}\cot x+c $
C) $ \frac{e^{2x}\tan x}{2}+c $
D) $ \frac{e^{2x}\cot x}{2}+c $
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_{{}}^{{}}{e^{2x}\frac{1+\sin 2x}{1+\cos 2x},dx}=\int_{{}}^{{}}{e^{2x}[ \frac{1}{1+\cos 2x}+\frac{\sin 2x}{1+\cos 2x} ],dx} $ $ =\int_{{}}^{{}}{e^{2x}[ \frac{{{\sec }^{2}}x}{2}+\tan x ]},dx $ $ =\frac{1}{2}\int_{{}}^{{}}{e^{2x}{{\sec }^{2}}x,dx}+\int_{{}}^{{}}{e^{2x}\tan x,dx} $ $ =\frac{e^{2x}\tan x}{2}-\int_{{}}^{{}}{\frac{e^{2x}{{\sec }^{2}}x}{2},dx}+\int_{{}}^{{}}{\frac{e^{2x}{{\sec }^{2}}x}{2},dx}+c $ $ =\frac{e^{2x}\tan x}{2}+c. $
BETA
