\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
