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

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

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integral of sin^5(x/3)

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

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

Solution steps

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

Apply u-substitution

: \int sin^{5} (u) \cdot 3 d u

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int - 1 + 2 v^{2} - v^{4} d v

= 3 \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

= 3 (- \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}

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

Substitute back

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

Simplify

3 (- cos (\frac{x}{3}) + \frac{2 cos ^{3} ( \frac{x}{3} )}{3} - \frac{cos ^{5} ( \frac{x}{3} )}{5}) : - 3 cos (\frac{x}{3}) + 2 cos^{3} (\frac{x}{3}) - \frac{3}{5} cos^{5} (\frac{x}{3})

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

Add a constant to the solution

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

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