What is the integral of sin^4(x/3) ?

The integral of sin^4(x/3) is 3(-1/4 sin^3(x/3)cos(x/3)+x/8-3/16 sin((2x)/3))+C

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integral of sin^4(x/3)

\int sin^{4} (\frac{x}{3}) d x

3 (- \frac{1}{4} sin^{3} (\frac{x}{3}) cos (\frac{x}{3}) + \frac{x}{8} - \frac{3}{16} sin (\frac{2 x}{3})) + C

Solution steps

\int sin^{4} (\frac{x}{3}) d x

Apply u-substitution

: \int sin^{4} (u) \cdot 3 d u

= \int sin^{4} (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^{4} (u) d u

Apply Integral Reduction

= 3 (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \int sin^{2} (u) d u)

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

= 3 (- \frac{cos ( u ) sin ^{3} ( u )}{4} + \frac{3}{4} \cdot \frac{1}{2} (u - \frac{1}{2} sin (2 u)))

Substitute back

u = \frac{x}{3}

= 3 (- \frac{cos ( \frac{x}{3} ) sin ^{3} ( \frac{x}{3} )}{4} + \frac{3}{4} \cdot \frac{1}{2} (\frac{x}{3} - \frac{1}{2} sin (2 \cdot \frac{x}{3})))

Simplify

3 (- \frac{cos ( \frac{x}{3} ) sin ^{3} ( \frac{x}{3} )}{4} + \frac{3}{4} \cdot \frac{1}{2} (\frac{x}{3} - \frac{1}{2} sin (2 \cdot \frac{x}{3}))) : 3 (- \frac{1}{4} sin^{3} (\frac{x}{3}) cos (\frac{x}{3}) + \frac{x}{8} - \frac{3}{16} sin (\frac{2 x}{3}))

= 3 (- \frac{1}{4} sin^{3} (\frac{x}{3}) cos (\frac{x}{3}) + \frac{x}{8} - \frac{3}{16} sin (\frac{2 x}{3}))

Add a constant to the solution

= 3 (- \frac{1}{4} sin^{3} (\frac{x}{3}) cos (\frac{x}{3}) + \frac{x}{8} - \frac{3}{16} sin (\frac{2 x}{3})) + C

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What is the integral of sin^4(x/3) ?
The integral of sin^4(x/3) is 3(-1/4 sin^3(x/3)cos(x/3)+x/8-3/16 sin((2x)/3))+C