What is the integral from 0 to pi of sin^4(at) ?

The integral from 0 to pi of sin^4(at) is (-1/4 sin^3(pia)cos(pia)+3/8 (pia-1/2 sin(2pia)))/a

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integral from 0 to pi of sin^4(at)

\int_{0}^{π} sin^{4} (a t) d t

\frac{- \frac{1}{4} sin ^{3} ( πa ) cos ( πa ) + \frac{3}{8} ( πa - \frac{1}{2} sin ( 2 πa ) )}{a}

Solution steps

\int_{0}^{π} sin^{4} (a t) d t

Apply u-substitution

: \int_{0}^{aπ} \frac{sin ^{4} ( u )}{a} d u

= \int_{0}^{aπ} \frac{sin ^{4} ( u )}{a} d u

Take the constant out:

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

= \frac{1}{a} \cdot \int_{0}^{aπ} sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{a} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{aπ} + \frac{3}{4} \cdot \int_{0}^{aπ} sin^{2} (u) d u)

\int_{0}^{aπ} sin^{2} (u) d u = \frac{1}{2} (πa - \frac{1}{2} sin (2 πa))

= \frac{1}{a} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{aπ} + \frac{3}{4} \cdot \frac{1}{2} (πa - \frac{1}{2} sin (2 πa)))

Simplify

\frac{1}{a} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{aπ} + \frac{3}{4} \cdot \frac{1}{2} (πa - \frac{1}{2} sin (2 πa))) : \frac{1}{a} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{πa} + \frac{3}{8} (πa - \frac{1}{2} sin (2 πa)))

= \frac{1}{a} (- [\frac{1}{4} sin^{3} (u) cos (u)]_{0}^{πa} + \frac{3}{8} (πa - \frac{1}{2} sin (2 πa)))

Compute the boundaries:

\frac{1}{4} sin^{3} (πa) cos (πa)

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

Simplify

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

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What is the integral from 0 to pi of sin^4(at) ?
The integral from 0 to pi of sin^4(at) is (-1/4 sin^3(pia)cos(pia)+3/8 (pia-1/2 sin(2pia)))/a