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

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

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integral of sin^3(2x-3)cos(2x-3)

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

\frac{1}{8} sin^{4} (2 x - 3) + C

Solution steps

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

Apply u-substitution

: \int sin^{3} (u) cos (u) \frac{1}{2} d u

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

Apply u-substitution

: \int v^{3} d v

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

Apply the Power Rule

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

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

Substitute back

= \frac{1}{2} \cdot \frac{sin ^{4} ( 2 x - 3 )}{4}

Simplify

\frac{1}{2} \cdot \frac{sin ^{4} ( 2 x - 3 )}{4} : \frac{1}{8} sin^{4} (2 x - 3)

= \frac{1}{8} sin^{4} (2 x - 3)

Add a constant to the solution

= \frac{1}{8} sin^{4} (2 x - 3) + C

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