What is the integral from 0 to pi of 8sin^4(x) ?

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

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integral from 0 to pi of 8sin^4(x)

\int_{0}^{π} 8 sin^{4} (x) d x

3 π

Decimal Notation

9.42477 \dots

Solution steps

\int_{0}^{π} 8 sin^{4} (x) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 8 \cdot \int_{0}^{π} sin^{4} (x) d x

Apply Integral Reduction

= 8 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \int_{0}^{π} sin^{2} (x) d x)

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

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

Simplify

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

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

Compute the boundaries:

0

= - 8 \cdot 0 + 3 π

Simplify

= 3 π

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