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

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

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

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

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

Solution steps

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

f (x) = x^{5} - 5 x^{3} + x - 1

4 x^{2} + 9 x - 24, 5 x^{2} + 21 x + 3

y^{^{'}} + \frac{y}{t} = 0

(e^{2})^{^{'}}

\frac{d}{d θ} (2 a sin (θ))

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