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

The integral of cos^5(3x)sin^3(3x) is 1/3 (-(cos^6(3x))/6+(cos^8(3x))/8)+C

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

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

\frac{1}{3} (- \frac{cos ^{6} ( 3 x )}{6} + \frac{cos ^{8} ( 3 x )}{8}) + C

Solution steps

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

Apply u-substitution

= \int cos^{5} (u) sin^{3} (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 cos^{5} (u) sin^{3} (u) d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int (1 - cos^{2} (u)) sin (u) cos^{5} (u) d u

Apply u-substitution

= \frac{1}{3} \cdot \int - v^{5} (1 - v^{2}) d v

Expand

- v^{5} (1 - v^{2}) : - v^{5} + v^{7}

= \frac{1}{3} \cdot \int - v^{5} + v^{7} 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^{5} d v + \int v^{7} d v)

\int v^{5} d v = \frac{v ^{6}}{6}

\int v^{7} d v = \frac{v ^{8}}{8}

= \frac{1}{3} (- \frac{v ^{6}}{6} + \frac{v ^{8}}{8})

Substitute back

= \frac{1}{3} (- \frac{cos ^{6} ( 3 x )}{6} + \frac{cos ^{8} ( 3 x )}{8})

Add a constant to the solution

= \frac{1}{3} (- \frac{cos ^{6} ( 3 x )}{6} + \frac{cos ^{8} ( 3 x )}{8}) + C

What is the integral of cos^5(3x)sin^3(3x) ?
The integral of cos^5(3x)sin^3(3x) is 1/3 (-(cos^6(3x))/6+(cos^8(3x))/8)+C