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

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

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

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

\frac{π - 2}{8 π}

Decimal

0.04542 \dots

Solution steps

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

Rewrite using trig identities

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

Take the constant out:

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

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int_{0}^{\frac{1}{4}} 1 d x = \frac{1}{4}

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

= \frac{1}{2} (\frac{1}{4} - \frac{1}{2 π})

Simplify

\frac{1}{2} (\frac{1}{4} - \frac{1}{2 π}) : \frac{π - 2}{8 π}

= \frac{π - 2}{8 π}

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\int_{π}^{0} sin^{4} (θ) d θ

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