MCQ
$\int_{}^{} {{e^x}\sin x(\sin x + 2\cos x)} \;dx = $
- ✓${e^x}{\sin ^2}x + c$
- B${e^x}\sin x + c$
- C${e^x}\sin 2x + c$
- DNone of these
✓
Answer
Correct option: A.
${e^x}{\sin ^2}x + c$
a
(a)$\int_{}^{} {{e^x}\sin x(\sin x + 2\cos x)dx} $
$ = \int_{}^{} {{e^x}{{\sin }^2}x\,dx} + \int_{}^{} {{e^x}2\sin x\,\cos xdx} $
$ = \int_{}^{} {{e^x}{{\sin }^2}x\,dx} + \int_{}^{} {{e^x}\sin 2x\,dx} $
$ = {e^x}{\sin ^2}x - \int_{}^{} {{e^x}\sin 2x\,dx} + \int_{}^{} {{e^x}\sin 2x\,dx\, + c} $
$ = {e^x}{\sin ^2}x + c.$
Aliter : $\int_{}^{} {{e^x}({{\sin }^2}x + \sin 2x)dx = {e^x}{{\sin }^2}x + c.} $
(a)$\int_{}^{} {{e^x}\sin x(\sin x + 2\cos x)dx} $
$ = \int_{}^{} {{e^x}{{\sin }^2}x\,dx} + \int_{}^{} {{e^x}2\sin x\,\cos xdx} $
$ = \int_{}^{} {{e^x}{{\sin }^2}x\,dx} + \int_{}^{} {{e^x}\sin 2x\,dx} $
$ = {e^x}{\sin ^2}x - \int_{}^{} {{e^x}\sin 2x\,dx} + \int_{}^{} {{e^x}\sin 2x\,dx\, + c} $
$ = {e^x}{\sin ^2}x + c.$
Aliter : $\int_{}^{} {{e^x}({{\sin }^2}x + \sin 2x)dx = {e^x}{{\sin }^2}x + c.} $
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