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

The integral of sin^2(2x)*cos(2x) is 1/12 (sin(6x)-3sin(2x)cos(4x))+C

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

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

\frac{1}{12} (sin (6 x) - 3 sin (2 x) cos (4 x)) + C

Solution steps

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

Rewrite using trig identities

= \int \frac{1}{2} cos (2 x) (1 - cos (4 x)) d x

Take the constant out:

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

= \frac{1}{2} \cdot \int cos (2 x) (1 - cos (4 x)) d x

Apply Integration By Parts

: \frac{1}{2} sin (2 x) (1 - cos (4 x)) - \int 2 sin (4 x) sin (2 x) d x

= \frac{1}{2} (\frac{1}{2} sin (2 x) (1 - cos (4 x)) - \int 2 sin (4 x) sin (2 x) d x)

\int 2 sin (4 x) sin (2 x) d x = - \frac{1}{6} sin (6 x) + \frac{1}{2} sin (2 x)

= \frac{1}{2} (\frac{1}{2} sin (2 x) (1 - cos (4 x)) - (- \frac{1}{6} sin (6 x) + \frac{1}{2} sin (2 x)))

Simplify

\frac{1}{2} (\frac{1}{2} sin (2 x) (1 - cos (4 x)) - (- \frac{1}{6} sin (6 x) + \frac{1}{2} sin (2 x))) : \frac{1}{12} (sin (6 x) - 3 sin (2 x) cos (4 x))

= \frac{1}{12} (sin (6 x) - 3 sin (2 x) cos (4 x))

Add a constant to the solution

= \frac{1}{12} (sin (6 x) - 3 sin (2 x) cos (4 x)) + C

f (x) = \frac{1}{x}, at x = 1.003

x^{5} e^{2 x}

\frac{\partial}{\partial x} (e^{- 3 π t} cos (4 x) sin (6 y))

y = in (x)

\int_{0}^{9} \frac{1}{( x - 1 ) ^{\frac{2}{3}}} d x

What is the integral of sin^2(2x)*cos(2x) ?
The integral of sin^2(2x)*cos(2x) is 1/12 (sin(6x)-3sin(2x)cos(4x))+C