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

The integral of sin^5(x/3)cos(x/3) is 1/2 sin^6(x/3)+C

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

integral

sin^{5} (\frac{x}{3}) cos (\frac{x}{3})

\frac{1}{2} sin^{6} (\frac{x}{3}) + C

Solution steps

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

Apply u-substitution

: \int \frac{3 5 u}{5} d u

= \int \frac{3 5 u}{5} d u

Take the constant out:

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

= \frac{3}{5} \cdot \int 5 u d u

Apply the Power Rule

: \frac{5}{6} u^{\frac{6}{5}}

= \frac{3}{5} \cdot \frac{5}{6} u^{\frac{6}{5}}

Substitute back

u = sin^{5} (\frac{x}{3})

= \frac{3}{5} \cdot \frac{5}{6} (sin^{5} (\frac{x}{3}))^{\frac{6}{5}}

Simplify

\frac{3}{5} \cdot \frac{5}{6} (sin^{5} (\frac{x}{3}))^{\frac{6}{5}} : \frac{1}{2} sin^{6} (\frac{x}{3})

= \frac{1}{2} sin^{6} (\frac{x}{3})

Add a constant to the solution

= \frac{1}{2} sin^{6} (\frac{x}{3}) + C

\frac{3 x ^{3}}{( 1 - x )}

\frac{1}{1} + x^{2} \frac{x ^{2}}{2}

\frac{- y}{( 1 + y ^{2} )}

\frac{1}{x} (arctan (\frac{1}{x}))

- 4 e^{3} x

What is the integral of sin^5(x/3)cos(x/3) ?
The integral of sin^5(x/3)cos(x/3) is 1/2 sin^6(x/3)+C