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

MCQ

$\int_{}^{} {{e^{2x}}( - \sin x + 2\cos x)\;dx = } $

A
${e^{2x}}\sin x + c$
B
$ - {e^{2x}}\sin x + c$
C
$ - {e^{2x}}\cos x + c$
✓
${e^{2x}}\cos x + c$

✓

Answer

Correct option: D.

${e^{2x}}\cos x + c$

d
(d)$\int_{}^{} {{e^{2x}}( - \sin x + 2\cos x)\,dx} $
$ = - \int_{}^{} {{e^{2x}}\sin x\,dx} + 2\int_{}^{} {{e^{2x}}\cos x\,dx} $
$ = {e^{2x}}\cos x - 2\int_{}^{} {{e^{2x}}\cos x\,dx + 2\int_{}^{} {{e^{2x}}\cos x\,dx + c} } $
$ = {e^{2x}}\cos x + c.$
Aliter : $\int_{}^{} {{e^{2x}}(2\cos x - \sin x)\,dx} = {e^{2x}}\cos x + c$
$\left\{ \because \,\,\,\int_{{}}^{{}}{{{e}^{kx}}\left\{ k\,f(x)+{f}'(x) \right\}dx={{e}^{kx}}f(x)+c} \right\}$

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