What is the integral of sin^2(u)cos^2(u) ?

The integral of sin^2(u)cos^2(u) is 1/8 (u-1/4 sin(4u))+C

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integral of sin^2(u)cos^2(u)

\int sin^{2} (u) cos^{2} (u) d u

\frac{1}{8} (u - \frac{1}{4} sin (4 u)) + C

Solution steps

\int sin^{2} (u) cos^{2} (u) d u

Rewrite using trig identities

= \int \frac{1 - cos ( 4 u )}{8} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{8} \cdot \int 1 - cos (4 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} (\int 1 d u - \int cos (4 u) d u)

\int 1 d u = u

\int cos (4 u) d u = \frac{1}{4} sin (4 u)

= \frac{1}{8} (u - \frac{1}{4} sin (4 u))

Add a constant to the solution

= \frac{1}{8} (u - \frac{1}{4} sin (4 u)) + C

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What is the integral of sin^2(u)cos^2(u) ?
The integral of sin^2(u)cos^2(u) is 1/8 (u-1/4 sin(4u))+C