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STD 12 - 7.1 inderfinite integral — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 7.1 inderfinite integral1 Mark
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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