_ ^ e^ 2x 1 + 2x 1 + 2x ;dx = - Vidyadip

MCQ

$\int_{}^{} {{e^{2x}}\frac{{1 + \sin 2x}}{{1 + \cos 2x}}} \;dx = $

A
${e^{2x}}\tan x + c$
B
${e^{2x}}\cot x + c$
✓
$\frac{{{e^{2x}}\tan x}}{2} + c$
D
$\frac{{{e^{2x}}\cot x}}{2} + c$

✓

Answer

Correct option: C.

$\frac{{{e^{2x}}\tan x}}{2} + c$

c
(c)$\int_{}^{} {{e^{2x}}\frac{{1 + \sin 2x}}{{1 + \cos 2x}}\,dx} = \int_{}^{} {{e^{2x}}\left[ {\frac{1}{{1 + \cos 2x}} + \frac{{\sin 2x}}{{1 + \cos 2x}}} \right]\,dx} $
$ = \int_{}^{} {{e^{2x}}\left[ {\frac{{{{\sec }^2}x}}{2} + \tan x} \right]} \,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.$

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