What is the integral of sin^4(w/2)cos^2(w/2) ?

The integral of sin^4(w/2)cos^2(w/2) is

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integral of sin^4(w/2)cos^2(w/2)

\int sin^{4} (\frac{w}{2}) cos^{2} (\frac{w}{2}) d w

2 (- \frac{1}{4} sin^{3} (\frac{w}{2}) cos (\frac{w}{2}) + \frac{3}{16} (w - sin (w)) + \frac{1}{6} sin^{5} (\frac{w}{2}) cos (\frac{w}{2}) - \frac{5}{96} (- 4 sin^{3} (\frac{w}{2}) cos (\frac{w}{2}) + 3 (w - sin (w)))) + C

Solution steps

\int sin^{4} (\frac{w}{2}) cos^{2} (\frac{w}{2}) d w

Apply u-substitution

: \int sin^{4} (u) cos^{2} (u) \cdot 2 d u

= \int sin^{4} (u) cos^{2} (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^{4} (u) cos^{2} (u) d u

Rewrite using trig identities

= 2 \cdot \int sin^{4} (u) (1 - sin^{2} (u)) d u

Expand

sin^{4} (u) (1 - sin^{2} (u)) : sin^{4} (u) - sin^{6} (u)

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 (\int sin^{4} (u) d u - \int sin^{6} (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))

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

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

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

Simplify

2 (- \frac{1}{4} sin^{3} (\frac{w}{2}) cos (\frac{w}{2}) + \frac{3}{8} (\frac{w}{2} - \frac{1}{2} sin (2 \cdot \frac{w}{2})) - (- \frac{cos ( \frac{w}{2} ) sin ^{5} ( \frac{w}{2} )}{6} + \frac{5}{6} (- \frac{1}{4} sin^{3} (\frac{w}{2}) cos (\frac{w}{2}) + \frac{3}{8} (\frac{w}{2} - \frac{1}{2} sin (2 \cdot \frac{w}{2}))))) : 2 (- \frac{1}{4} sin^{3} (\frac{w}{2}) cos (\frac{w}{2}) + \frac{3}{16} (w - sin (w)) + \frac{1}{6} sin^{5} (\frac{w}{2}) cos (\frac{w}{2}) - \frac{5}{96} (- 4 sin^{3} (\frac{w}{2}) cos (\frac{w}{2}) + 3 (w - sin (w))))

= 2 (- \frac{1}{4} sin^{3} (\frac{w}{2}) cos (\frac{w}{2}) + \frac{3}{16} (w - sin (w)) + \frac{1}{6} sin^{5} (\frac{w}{2}) cos (\frac{w}{2}) - \frac{5}{96} (- 4 sin^{3} (\frac{w}{2}) cos (\frac{w}{2}) + 3 (w - sin (w))))

Add a constant to the solution

= 2 (- \frac{1}{4} sin^{3} (\frac{w}{2}) cos (\frac{w}{2}) + \frac{3}{16} (w - sin (w)) + \frac{1}{6} sin^{5} (\frac{w}{2}) cos (\frac{w}{2}) - \frac{5}{96} (- 4 sin^{3} (\frac{w}{2}) cos (\frac{w}{2}) + 3 (w - sin (w)))) + C

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What is the integral of sin^4(w/2)cos^2(w/2) ?
The integral of sin^4(w/2)cos^2(w/2) is