What is the integral from 0 to 2pi of sqrt((1-cos(x))/2) ?

The integral from 0 to 2pi of sqrt((1-cos(x))/2) is 4

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integral from 0 to 2pi of sqrt((1-cos(x))/2)

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

4

Solution steps

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

\frac{1 - cos ( x )}{2} = \frac{1}{2} 1 - cos (x)

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

Rewrite using trig identities

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

Apply exponent rule:

a^{2} = ∣ a ∣

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

Eliminate Absolutes

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

Apply u-substitution

: \int_{0}^{π} sin (u) \cdot 2 d u

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

Take the constant out:

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

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

Use the common integral:

\int sin (u) d u = - cos (u)

= \frac{1}{2} 2 \cdot 2 [- cos (u)]_{0}^{π}

Simplify

\frac{1}{2} 2 \cdot 2 [- cos (u)]_{0}^{π} : 2 [- cos (u)]_{0}^{π}

= 2 [- cos (u)]_{0}^{π}

Compute the boundaries:

2

= 2 \cdot 2

Simplify

= 4

\int \frac{x ^{2} - 16}{x ^{4}} d x

n = 1 \sum \infty \frac{7}{n ( n + 1 )}

x \to 1 + lim (\frac{1}{x ^{1 - x}})

f (x) = 3 (x + 1)

\frac{d y}{d x} - y = x e^{x}

What is the integral from 0 to 2pi of sqrt((1-cos(x))/2) ?
The integral from 0 to 2pi of sqrt((1-cos(x))/2) is 4