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

$\int \cos ^3 x \cdot e^{\log (\sin x)} d x=$
  1. $\frac{-e^{\sin x}}{4}+c$
  2. $\frac{-\cos ^4 x}{4}+c$
  3. $\frac{-\sin ^4 x}{4}+c$
  4. $\frac{-e^{\sin x}}{4}+c$

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 $\cos \mathrm{x}=\mathrm{t}$ $\therefore I=\int-t^3 d t=\frac{-t^4}{4}+t=\frac{-\cos ^4 x}{4}+c$

Asked in: MHT CET 2021 (23 Sep Shift 1)

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