What is the integral from 0 to pi/3 of sin^6(x) ?

The integral from 0 to pi/3 of sin^6(x) is (20pi-27sqrt(3))/(192)

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integral from 0 to pi/3 of sin^6(x)

\int_{0}^{\frac{π}{3}} sin^{6} (x) d x

\frac{20 π - 27 3}{192}

Decimal

0.08367 \dots

Solution steps

\int_{0}^{\frac{π}{3}} sin^{6} (x) d x

Apply Integral Reduction

= - [\frac{cos ( x ) sin ^{5} ( x )}{6}]_{0}^{\frac{π}{3}} + \frac{5}{6} \cdot \int_{0}^{\frac{π}{3}} sin^{4} (x) d x

\int_{0}^{\frac{π}{3}} sin^{4} (x) d x = \frac{8 π - 9 3}{64}

= - [\frac{cos ( x ) sin ^{5} ( x )}{6}]_{0}^{\frac{π}{3}} + \frac{5}{6} \cdot \frac{8 π - 9 3}{64}

Simplify

- [\frac{cos ( x ) sin ^{5} ( x )}{6}]_{0}^{\frac{π}{3}} + \frac{5}{6} \cdot \frac{8 π - 9 3}{64} : - [\frac{1}{6} sin^{5} (x) cos (x)]_{0}^{\frac{π}{3}} + \frac{5 ( 8 π - 9 3 )}{384}

= - [\frac{1}{6} sin^{5} (x) cos (x)]_{0}^{\frac{π}{3}} + \frac{5 ( 8 π - 9 3 )}{384}

Compute the boundaries:

\frac{3 3}{128}

= - \frac{3 3}{128} + \frac{5 ( 8 π - 9 3 )}{384}

Simplify

= \frac{20 π - 27 3}{192}

What is the integral from 0 to pi/3 of sin^6(x) ?
The integral from 0 to pi/3 of sin^6(x) is (20pi-27sqrt(3))/(192)