_ ^ ( 2 + 2x 1 + 2x ) , , e^x dx = - Vidyadip

MCQ

$\int_{}^{} {\left( {\frac{{2 + \sin 2x}}{{1 + \cos 2x}}} \right)\,\,{e^x}dx = } $

A
${e^x}\cot x + c$
B
$ - {e^x}\cot x + c$
C
$ - {e^x}\tan x + c$
✓
${e^x}\tan x + c$

✓

Answer

Correct option: D.

${e^x}\tan x + c$

d
(d)$\int_{}^{} {\left( {\frac{{2 + \sin 2x}}{{1 + \cos 2x}}} \right){\rm{ }}{e^x}\,dx} = \int_{}^{} {\left( {\frac{{2{e^x}}}{{1 + \cos 2x}}} \right)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} $.

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