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

The integral of sin^5(3x)cos^3(3x) is 1/3 (-(cos^4(3x))/4+(cos^6(3x))/3-(cos^8(3x))/8)+C

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

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

\frac{1}{3} (- \frac{cos ^{4} ( 3 x )}{4} + \frac{cos ^{6} ( 3 x )}{3} - \frac{cos ^{8} ( 3 x )}{8}) + C

Solution steps

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

Apply u-substitution

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

Rewrite using trig identities

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

Apply u-substitution

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

Expand

- v^{3} (1 - v^{2})^{2} : - v^{3} + 2 v^{5} - v^{7}

= \frac{1}{3} \cdot \int - v^{3} + 2 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^{3} d v + \int 2 v^{5} d v - \int v^{7} d v)

\int v^{3} d v = \frac{v ^{4}}{4}

\int 2 v^{5} d v = \frac{v ^{6}}{3}

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

= \frac{1}{3} (- \frac{v ^{4}}{4} + \frac{v ^{6}}{3} - \frac{v ^{8}}{8})

Substitute back

= \frac{1}{3} (- \frac{cos ^{4} ( 3 x )}{4} + \frac{cos ^{6} ( 3 x )}{3} - \frac{cos ^{8} ( 3 x )}{8})

Add a constant to the solution

= \frac{1}{3} (- \frac{cos ^{4} ( 3 x )}{4} + \frac{cos ^{6} ( 3 x )}{3} - \frac{cos ^{8} ( 3 x )}{8}) + C

What is the integral of sin^5(3x)cos^3(3x) ?
The integral of sin^5(3x)cos^3(3x) is 1/3 (-(cos^4(3x))/4+(cos^6(3x))/3-(cos^8(3x))/8)+C