What is the integral from 0 to 1 of sin^4(1/2 pix) ?

The integral from 0 to 1 of sin^4(1/2 pix) is 3/8

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integral from 0 to 1 of sin^4(1/2 pix)

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

\frac{3}{8}

Decimal Notation

0.375

Solution steps

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

Multiply

\frac{1}{2} π x : \frac{π}{2} x

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Reduction

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

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

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

Simplify

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

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

Compute the boundaries:

0

= - \frac{2}{π} \cdot 0 + \frac{3}{8}

Simplify

= \frac{3}{8}

- \frac{0.625}{s + 4}

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What is the integral from 0 to 1 of sin^4(1/2 pix) ?
The integral from 0 to 1 of sin^4(1/2 pix) is 3/8