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

The integral from 0 to pi of sqrt(2-2cos(t)) is 4

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integral from 0 to pi of sqrt(2-2cos(t))

\int_{0}^{π} 2 - 2 cos (t) d t

4

Solution steps

\int_{0}^{π} 2 - 2 cos (t) d t

Apply u-substitution

: \int_{1}^{- 1} - \frac{2}{u + 1} d u

= \int_{1}^{- 1} - \frac{2}{u + 1} d u

\int_{a}^{b} f (x) d x = - \int_{b}^{a} f (x) d x, a < b

= - \int_{- 1}^{1} - \frac{2}{u + 1} d u

Take the constant out:

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

= - (- \int_{- 1}^{1} \frac{2}{u + 1} d u)

Apply Integration By Parts

: [\frac{2}{u + 1} u - \int - \frac{u}{2 ( u + 1 ) ^{\frac{3}{2}}} d u]_{- 1}^{1}

= - (- [\frac{2}{u + 1} u - \int - \frac{u}{2 ( u + 1 ) ^{\frac{3}{2}}} d u]_{- 1}^{1})

\int - \frac{u}{2 ( u + 1 ) ^{\frac{3}{2}}} d u = - \frac{2 ( u + 2 )}{u + 1}

= - (- [\frac{2}{u + 1} u - (- \frac{2 ( u + 2 )}{u + 1})]_{- 1}^{1})

Simplify

- (- [\frac{2}{u + 1} u - (- \frac{2 ( u + 2 )}{u + 1})]_{- 1}^{1}) : [22 1 + u]_{- 1}^{1}

= [22 1 + u]_{- 1}^{1}

Compute the boundaries:

4

= 4

\int_{0}^{2} \frac{x}{1 + 10 x ^{2}} d x

\int_{1}^{8} (\frac{1}{x}) d x

\int_{0}^{2} 4 - x d x

\int_{- 3}^{1} 8 d x

\int_{1}^{2} (4 x^{2}) d x

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