What is the integral of sin^7(3x)cos(3x) ?

The integral of sin^7(3x)cos(3x) is 1/24 sin^8(3x)+C

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integral of sin^7(3x)cos(3x)

\int sin^{7} (3 x) cos (3 x) d x

\frac{1}{24} sin^{8} (3 x) + C

Solution steps

\int sin^{7} (3 x) cos (3 x) d x

Apply u-substitution

: \int sin^{7} (u) cos (u) \frac{1}{3} d u

= \int sin^{7} (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^{7} (u) cos (u) d u

Apply u-substitution

: \int v^{7} d v

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

Apply the Power Rule

: \frac{v ^{8}}{8}

= \frac{1}{3} \cdot \frac{v ^{8}}{8}

Substitute back

= \frac{1}{3} \cdot \frac{sin ^{8} ( 3 x )}{8}

Simplify

\frac{1}{3} \cdot \frac{sin ^{8} ( 3 x )}{8} : \frac{1}{24} sin^{8} (3 x)

= \frac{1}{24} sin^{8} (3 x)

Add a constant to the solution

= \frac{1}{24} sin^{8} (3 x) + C

\frac{d}{d x} (\frac{1}{4} x - \frac{1}{10})

\int 5 x e^{7 x} d x

f (x) = 2^{60}

4 y^{^{''}} + 4 y^{^{'}} + y = 0, y (2) = - 6, y^{^{'}} (2) = 4

t^{6} e^{-, at}

What is the integral of sin^7(3x)cos(3x) ?
The integral of sin^7(3x)cos(3x) is 1/24 sin^8(3x)+C