What is the integral of sin^5(9x) ?

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

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integral of sin^5(9x)

\int sin^{5} (9 x) d x

\frac{1}{9} (- cos (9 x) + \frac{2 cos ^{3} ( 9 x )}{3} - \frac{cos ^{5} ( 9 x )}{5}) + C

Solution steps

\int sin^{5} (9 x) d x

Apply u-substitution

= \int sin^{5} (u) \frac{1}{9} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{9} \cdot \int sin^{5} (u) d u

Rewrite using trig identities

= \frac{1}{9} \cdot \int (1 - cos^{2} (u))^{2} sin (u) d u

Apply u-substitution

= \frac{1}{9} \cdot \int - 1 + 2 v^{2} - v^{4} 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}{9} (- \int 1 d v + \int 2 v^{2} d v - \int v^{4} d v)

\int 1 d v = v

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

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

= \frac{1}{9} (- v + \frac{2 v ^{3}}{3} - \frac{v ^{5}}{5})

Substitute back

= \frac{1}{9} (- cos (9 x) + \frac{2 cos ^{3} ( 9 x )}{3} - \frac{cos ^{5} ( 9 x )}{5})

Add a constant to the solution

= \frac{1}{9} (- cos (9 x) + \frac{2 cos ^{3} ( 9 x )}{3} - \frac{cos ^{5} ( 9 x )}{5}) + C

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