$\int \cos ^3 x e^{\log (\sin x)^2} d x=$

$\int \cos ^3 x e^{\log (\sin x)^2} d x=$
  1. $\frac{\sin ^3 x}{3}-\sin ^5 x+c$
  2. $\frac{\sin ^3 x}{3}-\frac{\sin ^5 x}{5}+c$
  3. $\frac{\sin ^3 x}{3}+\frac{\sin ^5 x}{5}+c$
  4. $\sin ^3 x+\sin ^5 x+c$

Solution

$\begin{aligned} & \text { Let } I=\int \cos ^3 x e^{\log (\sin x)^2} d x \\ & =\int \cos ^3 x(\sin x)^2 d x=\int \cos ^2 x(\sin x)^2 \cos x d x \\ & =\int \cos ^3 x(\sin x)^2 d x=\int \cos ^2 x(\sin x)^2 \cos x d x \end{aligned}$ Put $\sin x=t \Rightarrow \cos x d x=d t$ $\therefore \mathrm{I}=\int\left(\mathrm{t}^2-\mathrm{t}^4\right) \mathrm{dt}=\frac{\mathrm{t}^3}{3}-\frac{\mathrm{t}^5}{5}+\mathrm{c}=\frac{\sin ^3 \mathrm{x}}{3}-\frac{\sin ^5 \mathrm{x}}{5}+\mathrm{c}$

Asked in: MHT CET 2021 (24 Sep Shift 2)

Practice more Indefinite Integration questions on Aicharya