What is the integral of sin^4(1/2 x) ?

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

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integral of sin^4(1/2 x)

\int sin^{4} (\frac{1}{2} x) d x

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

Solution steps

\int sin^{4} (\frac{1}{2} x) d x

Simplify

\frac{1}{2} x : \frac{x}{2}

= \int sin^{4} (\frac{x}{2}) d x

Apply u-substitution

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

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

Apply Integral Reduction

= 2 (- \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))

= 2 (- \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}{2}

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

Simplify

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

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

Add a constant to the solution

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

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