#\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#
