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

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

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integral from 0 to pi of 2sin^4(x)

\int_{0}^{π} 2 sin^{4} (x) d x

\frac{3 π}{4}

Decimal

2.35619 \dots

Solution steps

\int_{0}^{π} 2 sin^{4} (x) d x

Take the constant out:

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

= 2 \cdot \int_{0}^{π} sin^{4} (x) d x

Apply Integral Reduction

= 2 (- [\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}

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

Simplify

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

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

Compute the boundaries:

0

= 2 (- 0 + \frac{3 π}{8})

Simplify

= \frac{3 π}{4}

\int \frac{x ^{3}}{x ^{2} + x + 1} d x

x^{2} y^{^{''}} - 7 x y^{^{'}} - 20 y = 0

y^{^{''}} + 9 y = 30 sin (t)

f (x) = e^{x} x

y = 8 cos (3 x), y = 8 sin (3 x), (0, \frac{π}{12})

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