Question

Answer 1

$$\begin{gathered} \mathrm{Given}\mathrm{integral}\mathrm{is}, \\ \int x^2{e}^{x^3}\cos \left(x^3\right)dx \\ \mathrm{Let}{x}^3=t \\ \Rightarrow 3x^{2}dx=dt \\ \Rightarrow x^{2}dx=\frac{dt}{3} \\ \mathrm{Integral}\mathrm{becomes}, \\ \frac{1}{3}\int e^t\cos tdt \\ =\frac{1}{3}I.....\left(1\right) \\ \mathrm{Where},\mathit{}I=\int e^t\cos tdt \\ I=\int e^t\cos tdt \\ \mathrm{Considering}\cos t\mathrm{as}\mathrm{first}\mathrm{and}{e}^t\mathrm{as}\mathrm{second}\mathrm{function} \\ I=\cos t{e}^t-\int -\sin t{e}^{t}dt \\ \Rightarrow I=e^t\cos t+\int \sin t{e}^{t}dt \\ \mathrm{Again}\mathrm{considering}\sin t\mathrm{as}\mathrm{first}\mathrm{and}{e}^t\mathrm{as}\mathrm{second}\mathrm{function} \\ I=e^t\cos t+\sin t{e}^t-\int \cos t{e}^{t}dt \\ \Rightarrow I=e^t\cos t+\sin t{e}^t-I \\ \Rightarrow 2I=e^t\left(\sin t+\cos t\right) \\ \Rightarrow I=\frac{e^t}{2}\left(\sin t+\cos t\right) \\ \therefore \int x^2{e}^{x^3}\cos \left(x^3\right)dx=\frac{1}{3}\left[\frac{e^t}{2}\left(\sin t+\cos t\right)\right]+\mathrm{C}\left[\mathrm{From}\left(1\right)\right] \\ =\frac{e^{x^3}}{6}\left(\sin x^3+\cos x^3\right)+\mathrm{C} \end{gathered}$$