_ ^ e^ - 2x 3x ;dx = - Vidyadip

MCQ

$\int_{}^{} {{e^{ - 2x}}\sin 3x\;dx = } $

A
$\frac{1}{{13}}{e^{ - 2x}}[\sin 3x + \cos 3x] + c$
B
$ - \frac{1}{{13}}{e^{ - 2x}}[\sin 3x + \cos 3x] + c$
C
$\frac{1}{{13}}{e^{ - 2x}}[2\sin 3x + 3\cos 3x] + c$
✓
$ - \frac{1}{{13}}{e^{ - 2x}}[2\sin 3x + 3\cos 3x] + c$

✓

Answer

Correct option: D.

$ - \frac{1}{{13}}{e^{ - 2x}}[2\sin 3x + 3\cos 3x] + c$

d
(d) Let $I = \int_{}^{} {{e^{ - 2x}}\sin 3x\,dx} $
$ = - \frac{{{e^{ - 2x}}\cos 3x}}{3} - \int_{}^{} {\frac{{2{e^{ - 2x}}\cos 3x}}{3}\,dx} $
$ = - \frac{{{e^{ - 2x}}\cos 3x}}{3} - \frac{2}{3}\left[ {\frac{{{e^{ - 2x}}\sin 3x}}{3} + \int_{}^{} {\frac{{2{e^{ - 2x}}\sin 3x}}{3}\,dx} } \right]$
$ \Rightarrow I = - \frac{{{e^{ - 2x}}\cos 3x}}{3} - \frac{{2{e^{ - 2x}}\sin 3x}}{9} - \frac{4}{9}I$
$ \Rightarrow \frac{{13}}{9}I = - {e^{ - 2x}}\left[ {\frac{{3\cos 3x + 2\sin 3x}}{9}} \right]$
Hence $I = - \frac{1}{{13}}{e^{ - 2x}}[3\cos 3x + 2\sin 3x]$.

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