What is the integral of sin^8(3x)cos^5(3x) ?

The integral of sin^8(3x)cos^5(3x) is 1/3 ((sin^9(3x))/9-(2sin^{11}(3x))/(11)+(sin^{13}(3x))/(13))+C

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integral of sin^8(3x)cos^5(3x)

\int sin^{8} (3 x) cos^{5} (3 x) d x

\frac{1}{3} (\frac{sin ^{9} ( 3 x )}{9} - \frac{2 sin ^{11} ( 3 x )}{11} + \frac{sin ^{13} ( 3 x )}{13}) + C

Solution steps

\int sin^{8} (3 x) cos^{5} (3 x) d x

Apply u-substitution

= \int sin^{8} (u) cos^{5} (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^{8} (u) cos^{5} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 - sin^{2} (u))^{2} cos (u) sin^{8} (u) d u

Apply u-substitution

= \frac{1}{3} \cdot \int v^{8} (1 - v^{2})^{2} d v

Expand

v^{8} (1 - v^{2})^{2} : v^{8} - 2 v^{10} + v^{12}

= \frac{1}{3} \cdot \int v^{8} - 2 v^{10} + v^{12} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int v^{8} d v - \int 2 v^{10} d v + \int v^{12} d v)

\int v^{8} d v = \frac{v ^{9}}{9}

\int 2 v^{10} d v = \frac{2 v ^{11}}{11}

\int v^{12} d v = \frac{v ^{13}}{13}

= \frac{1}{3} (\frac{v ^{9}}{9} - \frac{2 v ^{11}}{11} + \frac{v ^{13}}{13})

Substitute back

= \frac{1}{3} (\frac{sin ^{9} ( 3 x )}{9} - \frac{2 sin ^{11} ( 3 x )}{11} + \frac{sin ^{13} ( 3 x )}{13})

Add a constant to the solution

= \frac{1}{3} (\frac{sin ^{9} ( 3 x )}{9} - \frac{2 sin ^{11} ( 3 x )}{11} + \frac{sin ^{13} ( 3 x )}{13}) + C

What is the integral of sin^8(3x)cos^5(3x) ?
The integral of sin^8(3x)cos^5(3x) is 1/3 ((sin^9(3x))/9-(2sin^{11}(3x))/(11)+(sin^{13}(3x))/(13))+C