What is the integral of sin^7(3x)cos^2(3x) ?

The integral of sin^7(3x)cos^2(3x) is -1/9 cos^3(3x)+1/5 cos^5(3x)-1/7 cos^7(3x)+1/27 cos^9(3x)+C

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integral of sin^7(3x)cos^2(3x)

\int sin^{7} (3 x) cos^{2} (3 x) d x

- \frac{1}{9} cos^{3} (3 x) + \frac{1}{5} cos^{5} (3 x) - \frac{1}{7} cos^{7} (3 x) + \frac{1}{27} cos^{9} (3 x) + C

Solution steps

\int sin^{7} (3 x) cos^{2} (3 x) d x

Apply u-substitution

: \int sin^{7} (u) cos^{2} (u) \frac{1}{3} d u

= \int sin^{7} (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^{7} (u) cos^{2} (u) d u

Rewrite using trig identities

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

Apply u-substitution

: \int - v^{2} (1 - v^{2})^{3} d v

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

Expand

- v^{2} (1 - v^{2})^{3} : - v^{2} + 3 v^{4} - 3 v^{6} + v^{8}

= \frac{1}{3} \cdot \int - v^{2} + 3 v^{4} - 3 v^{6} + v^{8} 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 3 v^{4} d v - \int 3 v^{6} d v + \int v^{8} d v)

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

\int 3 v^{4} d v = \frac{3 v ^{5}}{5}

\int 3 v^{6} d v = \frac{3 v ^{7}}{7}

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

= \frac{1}{3} (- \frac{v ^{3}}{3} + \frac{3 v ^{5}}{5} - \frac{3 v ^{7}}{7} + \frac{v ^{9}}{9})

Substitute back

= \frac{1}{3} (- \frac{cos ^{3} ( 3 x )}{3} + \frac{3 cos ^{5} ( 3 x )}{5} - \frac{3 cos ^{7} ( 3 x )}{7} + \frac{cos ^{9} ( 3 x )}{9})

Simplify

\frac{1}{3} (- \frac{cos ^{3} ( 3 x )}{3} + \frac{3 cos ^{5} ( 3 x )}{5} - \frac{3 cos ^{7} ( 3 x )}{7} + \frac{cos ^{9} ( 3 x )}{9}) : - \frac{1}{9} cos^{3} (3 x) + \frac{1}{5} cos^{5} (3 x) - \frac{1}{7} cos^{7} (3 x) + \frac{1}{27} cos^{9} (3 x)

= - \frac{1}{9} cos^{3} (3 x) + \frac{1}{5} cos^{5} (3 x) - \frac{1}{7} cos^{7} (3 x) + \frac{1}{27} cos^{9} (3 x)

Add a constant to the solution

= - \frac{1}{9} cos^{3} (3 x) + \frac{1}{5} cos^{5} (3 x) - \frac{1}{7} cos^{7} (3 x) + \frac{1}{27} cos^{9} (3 x) + C

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What is the integral of sin^7(3x)cos^2(3x) ?
The integral of sin^7(3x)cos^2(3x) is -1/9 cos^3(3x)+1/5 cos^5(3x)-1/7 cos^7(3x)+1/27 cos^9(3x)+C