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

The integral from 0 to pi of 6sin^4(x) is (9pi)/4

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

\int_{0}^{π} 6 sin^{4} (x) d x

\frac{9 π}{4}

Decimal Notation

7.06858 \dots

Solution steps

\int_{0}^{π} 6 sin^{4} (x) d x

Take the constant out:

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

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

Apply Integral Reduction

= 6 (- [\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}

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

Simplify

6 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}) : 6 (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} + \frac{3 π}{8})

= 6 (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} + \frac{3 π}{8})

Compute the boundaries:

0

= 6 (- 0 + \frac{3 π}{8})

Simplify

= \frac{9 π}{4}

y = r^{2} - x^{2}, y = 0

\frac{\partial}{\partial x} ((ln (x))^{2})

x, at x = 144

\int \frac{x ^{3} + x ^{2} - x ^{5}}{x ^{6}} d x

\frac{d}{d x} (\frac{1}{4} (x^{4} + 8))

What is the integral from 0 to pi of 6sin^4(x) ?
The integral from 0 to pi of 6sin^4(x) is (9pi)/4