What is the integral of sin^6(3x)cos(3x) ?

The integral of sin^6(3x)cos(3x) is 1/21 sin^7(3x)+C

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integral of sin^6(3x)cos(3x)

\int sin^{6} (3 x) cos (3 x) d x

\frac{1}{21} sin^{7} (3 x) + C

Solution steps

\int sin^{6} (3 x) cos (3 x) d x

Apply u-substitution

= \int sin^{6} (u) cos (u) \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int sin^{6} (u) cos (u) d u

Apply u-substitution

= \frac{1}{3} \cdot \int v^{6} d v

Apply the Power Rule

= \frac{1}{3} \cdot \frac{v ^{7}}{7}

Substitute back

= \frac{1}{3} \cdot \frac{sin ^{7} ( 3 x )}{7}

Simplify

\frac{1}{3} \cdot \frac{sin ^{7} ( 3 x )}{7} : \frac{1}{21} sin^{7} (3 x)

= \frac{1}{21} sin^{7} (3 x)

Add a constant to the solution

= \frac{1}{21} sin^{7} (3 x) + C

What is the integral of sin^6(3x)cos(3x) ?
The integral of sin^6(3x)cos(3x) is 1/21 sin^7(3x)+C