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) 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

\frac{x ^{2}}{3 x + 7}

\frac{d y}{d x}, cos^{2} (x) + cos^{2} (y) = cos (2 x + 2 y)

\frac{d}{d x} (e^{\frac{x}{( x - 1 )}})

\frac{5}{x + 6}

\frac{d}{d x} (\frac{x ^{2} 3}{4})

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