Q75 / 150 · MHT CET 2021 (23 Sep Shift 1) · Mathematics

$\int \cos ^3 x \cdot e^{\log (\sin x)} d x=$

  • A
    $\frac{-e^{\sin x}}{4}+c$
  • Correct answer
    B
    $\frac{-\cos ^4 x}{4}+c$
  • C
    $\frac{-\sin ^4 x}{4}+c$
  • D
    $\frac{-e^{\sin x}}{4}+c$

Solution

Let $$ \begin{aligned} I & =\int \cos ^3 x \cdot e^{\log (\sin x)} d x \\ & =\int \cos ^3 x \cdot \sin x d x \end{aligned} $$ Put $\cos \mathrm{x}=\mathrm{t}$ $$ \therefore I=\int-t^3 d t=\frac{-t^4}{4}+t=\frac{-\cos ^4 x}{4}+c $$

Hence, the correct answer is option: 'B'.

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