What is the integral of sin^6(x/2) ?

The integral of sin^6(x/2) is 2(-1/6 sin^5(x/2)cos(x/2)+5/96 (-4sin^3(x/2)cos(x/2)+3(x-sin(x))))+C

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integral of sin^6(x/2)

\int sin^{6} (\frac{x}{2}) d x

2 (- \frac{1}{6} sin^{5} (\frac{x}{2}) cos (\frac{x}{2}) + \frac{5}{96} (- 4 sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) + 3 (x - sin (x)))) + C

Solution steps

\int sin^{6} (\frac{x}{2}) d x

Apply u-substitution

: \int sin^{6} (u) \cdot 2 d u

= \int sin^{6} (u) \cdot 2 d u

Take the constant out:

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

= 2 \cdot \int sin^{6} (u) d u

Apply Integral Reduction

= 2 (- \frac{cos ( u ) sin ^{5} ( u )}{6} + \frac{5}{6} \cdot \int sin^{4} (u) d u)

\int sin^{4} (u) d u = - \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u))

= 2 (- \frac{cos ( u ) sin ^{5} ( u )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (u) cos (u) + \frac{3}{8} (u - \frac{1}{2} sin (2 u))))

Substitute back

u = \frac{x}{2}

= 2 (- \frac{cos ( \frac{x}{2} ) sin ^{5} ( \frac{x}{2} )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) + \frac{3}{8} (\frac{x}{2} - \frac{1}{2} sin (2 \cdot \frac{x}{2}))))

Simplify

2 (- \frac{cos ( \frac{x}{2} ) sin ^{5} ( \frac{x}{2} )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) + \frac{3}{8} (\frac{x}{2} - \frac{1}{2} sin (2 \cdot \frac{x}{2})))) : 2 (- \frac{1}{6} sin^{5} (\frac{x}{2}) cos (\frac{x}{2}) + \frac{5}{96} (- 4 sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) + 3 (x - sin (x))))

= 2 (- \frac{1}{6} sin^{5} (\frac{x}{2}) cos (\frac{x}{2}) + \frac{5}{96} (- 4 sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) + 3 (x - sin (x))))

Add a constant to the solution

= 2 (- \frac{1}{6} sin^{5} (\frac{x}{2}) cos (\frac{x}{2}) + \frac{5}{96} (- 4 sin^{3} (\frac{x}{2}) cos (\frac{x}{2}) + 3 (x - sin (x)))) + C

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What is the integral of sin^6(x/2) ?
The integral of sin^6(x/2) is 2(-1/6 sin^5(x/2)cos(x/2)+5/96 (-4sin^3(x/2)cos(x/2)+3(x-sin(x))))+C