What is the integral from 0 to pi/2 of sin^6(t) ?

The integral from 0 to pi/2 of sin^6(t) is (5pi)/(32)

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integral from 0 to pi/2 of sin^6(t)

\int_{0}^{\frac{π}{2}} sin^{6} (t) d t

\frac{5 π}{32}

Decimal

0.49087 \dots

Solution steps

\int_{0}^{\frac{π}{2}} sin^{6} (t) d t

Apply Integral Reduction

= - [\frac{cos ( t ) sin ^{5} ( t )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \int_{0}^{\frac{π}{2}} sin^{4} (t) d t

\int_{0}^{\frac{π}{2}} sin^{4} (t) d t = \frac{3 π}{16}

= - [\frac{cos ( t ) sin ^{5} ( t )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \frac{3 π}{16}

Simplify

- [\frac{cos ( t ) sin ^{5} ( t )}{6}]_{0}^{\frac{π}{2}} + \frac{5}{6} \cdot \frac{3 π}{16} : - [\frac{1}{6} sin^{5} (t) cos (t)]_{0}^{\frac{π}{2}} + \frac{5 π}{32}

= - [\frac{1}{6} sin^{5} (t) cos (t)]_{0}^{\frac{π}{2}} + \frac{5 π}{32}

Compute the boundaries:

0

= - 0 + \frac{5 π}{32}

Simplify

= \frac{5 π}{32}

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What is the integral from 0 to pi/2 of sin^6(t) ?
The integral from 0 to pi/2 of sin^6(t) is (5pi)/(32)