What is the integral from 0 to pi of 12sin^4(x) ?

The integral from 0 to pi of 12sin^4(x) is (9pi)/2

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integral from 0 to pi of 12sin^4(x)

\int_{0}^{π} 12 sin^{4} (x) d x

\frac{9 π}{2}

Decimal Notation

14.13716 \dots

Solution steps

\int_{0}^{π} 12 sin^{4} (x) d x

Take the constant out:

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

= 12 \cdot \int_{0}^{π} sin^{4} (x) d x

Apply Integral Reduction

= 12 (- [\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}

= 12 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2})

Simplify

12 (- [\frac{cos ( x ) sin ^{3} ( x )}{4}]_{0}^{π} + \frac{3}{4} \cdot \frac{π}{2}) : 12 (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} + \frac{3 π}{8})

= 12 (- [\frac{1}{4} sin^{3} (x) cos (x)]_{0}^{π} + \frac{3 π}{8})

Compute the boundaries:

0

= 12 (- 0 + \frac{3 π}{8})

Simplify

= \frac{9 π}{2}

\int_{1}^{2} (7 x + x^{2}) d x

cos (2^{x})

(- 19 x y^{2} + 10 y) + (- 19 x^{2} y + 10 x) y^{^{'}} = 0

x^{2} y \frac{d y}{d x} = e^{y}

x \to 2 - lim (\frac{x - 2}{∣ x - 2 ∣})

What is the integral from 0 to pi of 12sin^4(x) ?
The integral from 0 to pi of 12sin^4(x) is (9pi)/2