Question

$\int {\sin }^{5/2}x{\cos }^{4}x$ dx

Answer 1

$\to \int {\sin }^{5/2}x{\cos }^{4}xdx$

Put $\sin x=t;\cos dx=dt$

$=\int t^{5/2}{\cos }^{3}x\cos xdx$

$=\int t^{5/2}(1-t^2{)}^{2}dt$

$=\int t^{5/2}(1+t^4-2t^2)dt$

$=\int (t^{5/2}+t^{13/2}-2t^{3/2})dt$

$=\frac{t^{7/2}}{7/2}+\frac{t^{15/2}}{15/2}-\frac{2t^{11/2}}{11/2}+C$

$=\frac{2}{7}{t}^{7/2}+\frac{2}{15}{t}^{15/2}-\frac{2}{11}{t}^{11/2}+C$

$=\frac{2}{7}(\sin x{)}^{7/2}+\frac{2}{15}(\sin x{)}^{15/2}-\frac{2}{11}(\sin x{)}^{11/2}$