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

The integral from 0 to pi of sin^4(2x) is (3pi)/8

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

\int_{0}^{π} sin^{4} (2 x) d x

\frac{3 π}{8}

Decimal

1.17809 \dots

Solution steps

\int_{0}^{π} sin^{4} (2 x) d x

Apply u-substitution

: \int_{0}^{2 π} sin^{4} (u) \frac{1}{2} d u

= \int_{0}^{2 π} sin^{4} (u) \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int_{0}^{2 π} sin^{4} (u) d u

Apply Integral Reduction

= \frac{1}{2} (- [\frac{cos ( u ) sin ^{3} ( u )}{4}]_{0}^{2 π} + \frac{3}{4} \cdot \int_{0}^{2 π} sin^{2} (u) d u)

\int_{0}^{2 π} sin^{2} (u) d u = π

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

Compute the boundaries:

0

= \frac{1}{2} (- 0 + \frac{3}{4} π)

Simplify

= \frac{3 π}{8}

x \to - \infty lim (x \cdot e^{x})

\frac{d}{d y} (sin^{2} (y))

\int_{- 10}^{10} π (100 - x^{2}) d x

\frac{\partial}{\partial y} (\frac{- 2 y}{( 2 x + 3 y ) ^{2}})

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

What is the integral from 0 to pi of sin^4(2x) ?
The integral from 0 to pi of sin^4(2x) is (3pi)/8