What is the integral of sin^4(piθ) ?

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

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integral of sin^4(piθ)

\int sin^{4} (π θ) d θ

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

Solution steps

\int sin^{4} (π θ) d θ

Apply u-substitution

: \int sin^{4} (u) \frac{1}{π} d u

= \int sin^{4} (u) \frac{1}{π} d u

Take the constant out:

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

= \frac{1}{π} \cdot \int sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{π} (- \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))

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

Substitute back

u = π θ

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

Simplify

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

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

Add a constant to the solution

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

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