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

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

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

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

\frac{1}{6} sin^{3} (2 x) + C

Solution steps

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

Apply u-substitution

: \int \frac{1}{2} sin^{2} (u) cos (u) d u

= \int \frac{1}{2} sin^{2} (u) cos (u) d u

Take the constant out:

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

= \frac{1}{2} \cdot \int sin^{2} (u) cos (u) d u

Apply u-substitution

: \int v^{2} d v

= \frac{1}{2} \cdot \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= \frac{1}{2} \cdot \frac{v ^{3}}{3}

Substitute back

= \frac{1}{2} \cdot \frac{sin ^{3} ( 2 x )}{3}

Simplify

\frac{1}{2} \cdot \frac{sin ^{3} ( 2 x )}{3} : \frac{1}{6} sin^{3} (2 x)

= \frac{1}{6} sin^{3} (2 x)

Add a constant to the solution

= \frac{1}{6} sin^{3} (2 x) + C

- 4 sin (4 t)

\frac{d}{d θ} (6 sin (θ) - 8 cos (θ))

\int 2 x^{- 2} d x

ln (4 z + 1)

x \to 1 lim (3 x^{3} - 4 x^{2} + x + 5)

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