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

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

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

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

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

Solution steps

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

\frac{d}{d x} (6 x^{5} e^{6 x})

\frac{d}{d x} (\frac{1 - 2 x}{6 + x})

t an g e n t f (x) = 4 x - 3 x

\frac{d}{d x} (a x^{n})

k = 1 \sum \infty \frac{k !}{2 ^{k}}

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