What is the integral from 0 to pi of 18sin^4(x) ?

The integral from 0 to pi of 18sin^4(x) is (27pi)/4

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integral from 0 to pi of 18sin^4(x)

\int_{0}^{π} 18 sin^{4} (x) d x

\frac{27 π}{4}

Decimal Notation

21.20575 \dots

Solution steps

\int_{0}^{π} 18 sin^{4} (x) d x

Take the constant out:

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

= 18 \cdot \int_{0}^{π} sin^{4} (x) d x

Apply Integral Reduction

= 18 (- [\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}

= 18 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2})

Simplify

18 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}) : 18 (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} + \frac{3 π}{8})

= 18 (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} + \frac{3 π}{8})

Compute the boundaries:

0

= 18 (- 0 + \frac{3 π}{8})

Simplify

= \frac{27 π}{4}

x y^{^{''}} + y^{^{'}} = 28 x

\frac{d}{d x} (ln (1 - 3 x^{2}))

\frac{d}{d x} (\frac{x ^{2} + 5 x + 6}{x ^{2} - 4})

\frac{\partial}{\partial y} (\frac{x ^{2}}{( x - y ) ^{2}})

\int - \frac{1}{x} d x

What is the integral from 0 to pi of 18sin^4(x) ?
The integral from 0 to pi of 18sin^4(x) is (27pi)/4