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

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

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

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

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

Solution steps

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

Rewrite using trig identities

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

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 (8 x) d x

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 x - \int cos (8 x) d x)

\int 1 d x = x

\int cos (8 x) d x = \frac{1}{8} sin (8 x)

= \frac{1}{8} (x - \frac{1}{8} sin (8 x))

Add a constant to the solution

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

\int_{0}^{\frac{π}{12}} \frac{sin ( 6 x )}{1 + cos ^{2} ( 6 x )} d x

3 y^{^{''}} - 2 y^{^{'}} + 4 y = 0

\frac{d}{d x} (2 sin (x^{2} - 1))

\int \frac{1}{- x ^{2} + 4 x + 12} d x

n = 1 \sum \infty \frac{8}{n ^{5}}

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