Question

$\int {\sin }^{3}x{\cos }^{4}xdx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int {\sin }^{3}x·{\cos }^{4}xdx \\ =\int {\sin }^{2}x·\sin x·{\cos }^{4}xdx \\ =\int \left(1-{\cos }^{2}x\right)·{\cos }^{4}x·\sin xdx \\ =\int \left({\cos }^{4}x-{\cos }^{6}x\right)·\sin xdx \\ \mathrm{Putting}\cos x=t \\ \Rightarrow -\sin xdx=dt \\ \Rightarrow \sin xdx=-dt \\ \therefore I=-\int \left(t^4-t^6\right)dt \\ =\int \left(t^6-t^4\right)dt \\ =\frac{t^7}{7}-\frac{t^5}{5}+\mathrm{C} \\ =\frac{{\cos }^{7}x}{7}-\frac{{\cos }^{5}x}{5}+\mathrm{C}\left[\because \mathit{}t=\cos \mathit{}x\right] \end{gathered}$$