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

The integral from 0 to pi of sin^4(t) is (3pi)/8

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

\int_{0}^{π} sin^{4} (t) d t

\frac{3 π}{8}

Decimal Notation

1.17809 \dots

Solution steps

\int_{0}^{π} sin^{4} (t) d t

Apply Integral Reduction

= - [\frac{cos ( t ) sin ^{3} ( t )}{4}]_{0}^{π} + \frac{3}{4} \cdot \int_{0}^{π} sin^{2} (t) d t

\int_{0}^{π} sin^{2} (t) d t = \frac{π}{2}

= - [\frac{cos ( t ) sin ^{3} ( t )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}

Simplify

- [\frac{cos ( t ) sin ^{3} ( t )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2} : - [\frac{1}{4} sin^{3} (t) cos (t)]_{0}^{π} + \frac{3 π}{8}

= - [\frac{1}{4} sin^{3} (t) cos (t)]_{0}^{π} + \frac{3 π}{8}

Compute the boundaries:

0

= - 0 + \frac{3 π}{8}

Simplify

= \frac{3 π}{8}

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What is the integral from 0 to pi of sin^4(t) ?
The integral from 0 to pi of sin^4(t) is (3pi)/8