What is the integral of sin^5(3x)*cos^2(3x) ?

The integral of sin^5(3x)*cos^2(3x) is 1/3 (-(cos^3(3x))/3+(2cos^5(3x))/5-(cos^7(3x))/7)+C

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integral of sin^5(3x)*cos^2(3x)

\int sin^{5} (3 x) \cdot cos^{2} (3 x) d x

\frac{1}{3} (- \frac{cos ^{3} ( 3 x )}{3} + \frac{2 cos ^{5} ( 3 x )}{5} - \frac{cos ^{7} ( 3 x )}{7}) + C

Solution steps

\int sin^{5} (3 x) cos^{2} (3 x) d x

Apply u-substitution

= \int sin^{5} (u) cos^{2} (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^{2} (u) d u

Rewrite using trig identities

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

Apply u-substitution

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

Expand

- v^{2} (1 - v^{2})^{2} : - v^{2} + 2 v^{4} - v^{6}

= \frac{1}{3} \cdot \int - v^{2} + 2 v^{4} - v^{6} 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^{2} d v + \int 2 v^{4} d v - \int v^{6} d v)

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

\int 2 v^{4} d v = \frac{2 v ^{5}}{5}

\int v^{6} d v = \frac{v ^{7}}{7}

= \frac{1}{3} (- \frac{v ^{3}}{3} + \frac{2 v ^{5}}{5} - \frac{v ^{7}}{7})

Substitute back

= \frac{1}{3} (- \frac{cos ^{3} ( 3 x )}{3} + \frac{2 cos ^{5} ( 3 x )}{5} - \frac{cos ^{7} ( 3 x )}{7})

Add a constant to the solution

= \frac{1}{3} (- \frac{cos ^{3} ( 3 x )}{3} + \frac{2 cos ^{5} ( 3 x )}{5} - \frac{cos ^{7} ( 3 x )}{7}) + C

What is the integral of sin^5(3x)*cos^2(3x) ?
The integral of sin^5(3x)*cos^2(3x) is 1/3 (-(cos^3(3x))/3+(2cos^5(3x))/5-(cos^7(3x))/7)+C