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

The integral of cos^2(4x)sin^2(4x) is 1/8 (x-1/16 sin(16x))+C

integral of cos^2(4x)sin^2(4x)

\int cos^{2} (4 x) sin^{2} (4 x) d x

\frac{1}{8} (x - \frac{1}{16} sin (16 x)) + C

Solution steps

\int cos^{2} (4 x) sin^{2} (4 x) d x

Rewrite using trig identities

= \int \frac{1}{8} (1 - cos (16 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 (16 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 (16 x) d x)

\int 1 d x = x

\int cos (16 x) d x = \frac{1}{16} sin (16 x)

= \frac{1}{8} (x - \frac{1}{16} sin (16 x))

Add a constant to the solution

= \frac{1}{8} (x - \frac{1}{16} sin (16 x)) + C

\int (4 x^{4} + 3 x^{2} + 2 x + 1 + x^{- 5}) d x

\int (\frac{1 - 2 x ^{3}}{x ^{3}}) d x

\int \frac{2}{4} x^{3} + x^{- 2} + 2 x d x

\int x^{5} a x^{6} - b d x

\int sin (\frac{t}{3}) d t

What is the integral of cos^2(4x)sin^2(4x) ?
The integral of cos^2(4x)sin^2(4x) is 1/8 (x-1/16 sin(16x))+C