#\int \sin^2\(x\)cos^4\(x\)dx#
Applying integral reduction,
#int sin^2(x) cos^n (x) dx# = #((sin^3 (x) cos^(n-1 )(x)) / (2+n))# #+((n-1) /(2+n))# #int sin^2 (x) cos^(n-2) (x)dx#
so,
#\int \sin ^2\(x\)\cos ^4\(x\)dx#
#=\frac{\sin ^3\(x\)\cos ^3\(x\)}{6}+\frac{3}{6}\int \cos ^2\(x\)\sin ^2\(x\)dx#
#=\frac{\sin ^3\(x\)\cos ^3\(x\)}{6}+\frac{3}{6}\int \cos ^2\(x\)\sin ^2\(x\)dx#
We know,
#\int \cos ^2\(x\)\sin ^2\(x\)dx=\frac{1}{8}\(x-\frac{1}{4}\sin \(4x\)\)#
Then,
#=\frac{\sin ^3\(x\)\cos ^3\(x\)}{6}+\frac{3}{6}\frac{1}{8}\(x-\frac{1}{4}\sin \(4x\)\)#
Simplifying,
#=\frac{1}{16}\(x-\frac{1}{4}\sin \(4x\)\)+\frac{1}{6}\sin ^3\(x\)\cos ^3\(x\)#
Adding constant to the solution,
#=\frac{1}{16}\(x-\frac{1}{4}\sin \(4x\)\)+\frac{1}{6}\sin ^3\(x\)\cos ^3\(x\)+C#
