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

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

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

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

\frac{2 π - 4}{8 2}

Decimal Notation

0.03914 \dots

Solution steps

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

Rewrite using trig identities

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

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

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

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

Simplify

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

= \frac{2 π - 4}{8 2}

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\int_{\frac{- π}{4}}^{\frac{π}{4}} cos (x) d x

\int_{0}^{\frac{π}{4}} tan^{3} (θ) d θ

\int_{0}^{2} 15 e^{3 x} d x

\int_{0}^{π} - sin (\frac{n x}{2}) d x

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