Question

Answer 1

$$\begin{gathered} \int x^3\cos x^{2}dx \\ \mathrm{Let}{x}^2=t \\ \Rightarrow 2x=\frac{dt}{dx} \\ \Rightarrow dx=\frac{dt}{2x} \\ =\frac{1}{2}\left[\int t\cos tdt\right] \\ \mathrm{Taking}t\mathrm{as}\mathrm{the}\mathrm{first}\mathrm{function}\mathrm{and}\cos t\mathrm{as}\mathrm{the}\mathrm{second}\mathrm{function}. \\ =\frac{1}{2}\left[t\sin t-\int \sin tdt\right] \\ =\frac{1}{2}\left[t\sin t+\cos t\right]...\left(1\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}t\mathrm{in}\mathrm{eq}\left(1\right) \\ =\frac{x^2\sin x^2}{2}+\frac{\cos x^2}{2}+c \end{gathered}$$