What is the integral of sin^3(2ax)cos(2ax) ?

The integral of sin^3(2ax)cos(2ax) is 1/(8a)sin^4(2ax)+C

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integral of sin^3(2ax)cos(2ax)

\int sin^{3} (2 a x) cos (2 a x) d x

\frac{1}{8 a} sin^{4} (2 a x) + C

Solution steps

\int sin^{3} (2 a x) cos (2 a x) d x

Apply u-substitution

: \int \frac{sin ^{3} ( u ) cos ( u )}{2 a} d u

= \int \frac{sin ^{3} ( u ) cos ( u )}{2 a} d u

Take the constant out:

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

= \frac{1}{2 a} \cdot \int sin^{3} (u) cos (u) d u

Apply u-substitution

: \int v^{3} d v

= \frac{1}{2 a} \cdot \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

= \frac{1}{2 a} \cdot \frac{v ^{4}}{4}

Substitute back

= \frac{1}{2 a} \cdot \frac{sin ^{4} ( 2 a x )}{4}

Simplify

\frac{1}{2 a} \cdot \frac{sin ^{4} ( 2 a x )}{4} : \frac{1}{8 a} sin^{4} (2 a x)

= \frac{1}{8 a} sin^{4} (2 a x)

Add a constant to the solution

= \frac{1}{8 a} sin^{4} (2 a x) + C

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

\int \frac{ln ( x + 1 )}{x ^{2}} d x

\frac{d y}{d t} = y^{3} - y

\int \frac{1}{x 7 x - 1} d x

\frac{d}{d x} (e^{c o s (2 x)})

What is the integral of sin^3(2ax)cos(2ax) ?
The integral of sin^3(2ax)cos(2ax) is 1/(8a)sin^4(2ax)+C