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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int v^{2} (1 - v^{2})^{2} d v

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

Add a constant to the solution

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

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