What is the integral of int sin^2(x)cos^4(x) dx?

1 Answer
Jun 5, 2016

\frac{1}{16}\(x-\frac{1}{4}\sin \(4x\))+\frac{1}{6}\sin ^3\(x\)\cos ^3\(x)+C

Explanation:

\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