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

The integral from 0 to pi of 4sin^4(x) is (3pi)/2

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

\int_{0}^{π} 4 sin^{4} (x) d x

\frac{3 π}{2}

Solution steps

\int_{0}^{π} 4 sin^{4} (x) d x

Take the constant out:

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

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

Apply Integral Reduction

= 4 (- [\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}

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

Simplify

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

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

Compute the boundaries:

0

= 4 (- 0 + \frac{3 π}{8})

Simplify

= \frac{3 π}{2}

\frac{d}{d x} (x^{2} (x + 2)^{3})

x^{2} - 8 x

\int x 6 - x^{2} d x

f (x) = 4 x^{2} + 3 x

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (9 x)

What is the integral from 0 to pi of 4sin^4(x) ?
The integral from 0 to pi of 4sin^4(x) is (3pi)/2