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

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

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

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

\frac{5 π}{32}

Decimal Notation

0.49087 \dots

Solution steps

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

Simplify

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

Apply Integral Reduction

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

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

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

Simplify

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

= - [\frac{1}{6} sin^{5} (x) cos (x)]_{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(x))^6 ?
The integral from 0 to pi/2 of (sin(x))^6 is (5pi)/(32)