Integral Calculus Question 338
Question: $ \int_{{}}^{{}}{( \frac{2+\sin 2x}{1+\cos 2x} )e^{x}dx=} $
[AISSE 1982]
Options:
A) $ e^{x}\cot x+c $
B) $ -e^{x}\cot x+c $
C) $ -e^{x}\tan x+c $
D) $ e^{x}\tan x+c $
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Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{( \frac{2+\sin 2x}{1+\cos 2x} )e^{x},dx}=\int_{{}}^{{}}{( \frac{2e^{x}}{1+\cos 2x} )dx}+\int_{{}}^{{}}{\frac{e^{x}\sin 2x}{1+\cos 2x}dx} $ $ =\int_{{}}^{{}}{e^{x}{{\sec }^{2}}x,dx}+\int_{{}}^{{}}{e^{x}\tan x,dx=e^{x}\tan x+c} $ .
BETA
