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

The integral from 0 to 1 of sqrt(2-2cos(t)) is 2(-2/(sec(1/2))+2)

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integral from 0 to 1 of sqrt(2-2cos(t))

\int_{0}^{1} 2 - 2 cos (t) d t

2 (- \frac{2}{sec ( \frac{1}{2} )} + 2)

Decimal Notation

0.48966 \dots

Solution steps

\int_{0}^{1} 2 - 2 cos (t) d t

Apply u-substitution

: \int_{0}^{t a n (\frac{1}{2})} \frac{4 u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

= \int_{0}^{t a n (\frac{1}{2})} \frac{4 u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

Take the constant out:

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

= 4 \cdot \int_{0}^{t a n (\frac{1}{2})} \frac{u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

Apply u-substitution

: \int_{1}^{1 + t a n^{2} (\frac{1}{2})} \frac{1}{2 v v} d v

= 4 \cdot \int_{1}^{1 + t a n^{2} (\frac{1}{2})} \frac{1}{2 v v} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int_{1}^{1 + t a n^{2} (\frac{1}{2})} \frac{1}{v v} d v

Simplify

v v : v^{\frac{3}{2}}

= 4 \cdot \frac{1}{2} \cdot \int_{1}^{1 + t a n^{2} (\frac{1}{2})} \frac{1}{v ^{\frac{3}{2}}} d v

Apply the Power Rule

: [- \frac{2}{v}]_{1}^{1 + t a n^{2} (\frac{1}{2})}

= 4 \cdot \frac{1}{2} [- \frac{2}{v}]_{1}^{1 + t a n^{2} (\frac{1}{2})}

Simplify

4 \cdot \frac{1}{2} [- \frac{2}{v}]_{1}^{1 + t a n^{2} (\frac{1}{2})} : 2 [- \frac{2}{v}]_{1}^{s e c^{2} (\frac{1}{2})}

= 2 [- \frac{2}{v}]_{1}^{s e c^{2} (\frac{1}{2})}

Compute the boundaries:

- \frac{2}{sec ( \frac{1}{2} )} + 2

= 2 (- \frac{2}{sec ( \frac{1}{2} )} + 2)

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What is the integral from 0 to 1 of sqrt(2-2cos(t)) ?
The integral from 0 to 1 of sqrt(2-2cos(t)) is 2(-2/(sec(1/2))+2)