_ ^ e^ 2x ( 4x - 2 1 - 4x ) ;dx = - Vidyadip

MCQ

$\int_{}^{} {{e^{2x}}\left( {\frac{{\sin 4x - 2}}{{1 - \cos 4x}}} \right)\;dx = } $

✓
$\frac{1}{2}{e^{2x}}\cot 2x + c$
B
$ - \frac{1}{2}{e^{2x}}\cot 2x + c$
C
$ - 2{e^{2x}}\cot 2x + c$
D
$2{e^{2x}}\cot 2x + c$

✓

Answer

Correct option: A.

$\frac{1}{2}{e^{2x}}\cot 2x + c$

a
(a)$\int_{}^{} {{e^{2x}}\left( {\frac{{\sin 4x - 2}}{{1 - \cos 4x}}} \right)} \,dx = \int_{}^{} {\frac{{{e^{2x}}\sin 4x}}{{1 - \cos 4x}}dx - 2\int_{}^{} {\frac{{{e^{2x}}}}{{1 - \cos 4x}}dx} } $
$ = \int_{}^{} {{e^{2x}}\cot 2x\,dx} - \int_{}^{} {{e^{2x}}{\rm{cose}}{{\rm{c}}^{\rm{2}}}2x\,dx} $
$ = \frac{{{e^{2x}}\cot 2x}}{2} + \int_{}^{} {2\frac{{{e^{2x}}}}{2}{\rm{cose}}{{\rm{c}}^{\rm{2}}}2x\,dx} - \int_{}^{} {{e^{2x}}{\rm{cose}}{{\rm{c}}^{\rm{2}}}x\,dx} $
$ = \frac{1}{2}({e^{2x}}\cot 2x) + c.$

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