$\int \frac{\sin 2 x\left(1-\frac{3}{2} \cos x\right)}{\mathrm{e}^{\sin ^2 x+\cos ^3 x}} \mathrm{~d} x=$

$\int \frac{\sin 2 x\left(1-\frac{3}{2} \cos x\right)}{\mathrm{e}^{\sin ^2 x+\cos ^3 x}} \mathrm{~d} x=$
  1. $\mathrm{e}^{\sin ^2 x+\cos ^3 x}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  2. $-\mathrm{e}^{-\left(\sin ^2 x+\cos ^3 x\right)}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  3. $\mathrm{e}^{-\left(\sin ^2 x+\cos ^3 x\right)^2}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.
  4. $\mathrm{e}^{\sin ^2 x+\cos x}+\mathrm{c}$, where $\mathrm{c}$ is a constant of integration.

Solution

$\begin{aligned} & \text { Put } \sin ^2 x+\cos ^3 x=\mathrm{t} \\ & \Rightarrow\left(2 \sin x \cos x-3 \cos ^2 x \sin x\right) \mathrm{d} x=\mathrm{dt} \\ & \Rightarrow\left(\sin 2 x-\frac{3}{2} \sin 2 x \cos x\right) \mathrm{d} x=\mathrm{dt} \\ & \Rightarrow \sin 2 x\left(1-\frac{3}{2} \cos x\right) \mathrm{d} x=\mathrm{dt} \\ \therefore \quad & \int \frac{\sin 2 x\left(1-\frac{3}{2} \cos x\right)}{\mathrm{d}} \mathrm{d} x \\ & =\int \frac{1}{\mathrm{e}^{\mathrm{t}}} \mathrm{dt}^2 x+\cos ^3 x \\ & =\int \mathrm{e}^{-\mathrm{t}} \mathrm{dt} \\ & =-\mathrm{e}^{-\mathrm{t}}+\mathrm{c} \\ & =-\mathrm{e}^{-\left(\sin ^2 x+\cos ^3 x\right)}+\mathrm{c}\end{aligned}$

Asked in: MHT CET 2023 (14 May Shift 1)

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