What is the integral from 0 to pi/6 of sin^4(x) ?

The integral from 0 to pi/6 of sin^4(x) is (4pi-7sqrt(3))/(64)

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integral from 0 to pi/6 of sin^4(x)

\int_{0}^{\frac{π}{6}} sin^{4} (x) d x

\frac{4 π - 7 3}{64}

Decimal Notation

0.00690 \dots

Solution steps

\int_{0}^{\frac{π}{6}} sin^{4} (x) d x

Apply Integral Reduction

= - [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{\frac{π}{6}} + \frac{3}{4} \cdot \int_{0}^{\frac{π}{6}} sin^{2} (x) d x

\int_{0}^{\frac{π}{6}} sin^{2} (x) d x = \frac{2 π - 3 3}{24}

= - [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{\frac{π}{6}} + \frac{3}{4} \cdot \frac{2 π - 3 3}{24}

Simplify

- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{\frac{π}{6}} + \frac{3}{4} \cdot \frac{2 π - 3 3}{24} : - [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{\frac{π}{6}} + \frac{2 π - 3 3}{32}

= - [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{\frac{π}{6}} + \frac{2 π - 3 3}{32}

Compute the boundaries:

\frac{3}{64}

= - \frac{3}{64} + \frac{2 π - 3 3}{32}

Simplify

= \frac{4 π - 7 3}{64}

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What is the integral from 0 to pi/6 of sin^4(x) ?
The integral from 0 to pi/6 of sin^4(x) is (4pi-7sqrt(3))/(64)