MCQ
$\int_{}^{} {{{\cos }^3}{\kern 1pt} x\;{e^{\log (\sin x)}}} \;dx$ is equal to
  • A
    $ - \frac{{{{\sin }^4}x}}{4} + c$
  • $ - \frac{{{{\cos }^4}x}}{4} + c$
  • C
    $\frac{{{e^{\sin x}}}}{4} + c$
  • D
    None of these

Answer

Correct option: B.
$ - \frac{{{{\cos }^4}x}}{4} + c$
b
(b)$\int_{}^{} {{{\cos }^3}x\,\,{e^{\log \sin x}}dx} = \int_{}^{} {{{\cos }^3}x\sin x\,dx} $
$ = - \int_{}^{} {{t^3}dt} = - \frac{{{t^4}}}{4} + c = - \frac{{{{\cos }^4}x}}{4} + c$ $\{ {\rm{Putting}}\,t = \cos x]$.

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