What is the integral of sqrt(2-2cos(x)) ?

The integral of sqrt(2-2cos(x)) is -4cos(x/2)+C

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integral of sqrt(2-2cos(x))

\int 2 - 2 cos (x) d x

- 4 cos (\frac{x}{2}) + C

Solution steps

\int 2 - 2 cos (x) d x

Apply u-substitution

: \int \frac{4 u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

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

Apply u-substitution

: \int \frac{1}{2 v v} d v

= 4 \cdot \int \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 \frac{1}{v v} d v

Simplify

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{\frac{3}{2}}} d v

Apply the Power Rule

: - \frac{2}{v}

= 4 \cdot \frac{1}{2} (- \frac{2}{v})

Substitute back

= 4 \cdot \frac{1}{2} - \frac{2}{1 + tan ^{2} ( \frac{x}{2} )}

Simplify

4 \cdot \frac{1}{2} - \frac{2}{1 + tan ^{2} ( \frac{x}{2} )} : - 4 cos (\frac{x}{2})

= - 4 cos (\frac{x}{2})

Add a constant to the solution

= - 4 cos (\frac{x}{2}) + C

4 x^{\frac{5}{4}} + 2 x^{\frac{3}{2}} + 4 x

\frac{3}{z ^{2}}

\int \frac{2 x - 1}{5 x ^{2} - x + 3} d x

\int \frac{e ^{4 x}}{e ^{x}} d x

\int \frac{5 t ^{3} - 2 t ^{5}}{3 t ^{2}} d t

What is the integral of sqrt(2-2cos(x)) ?
The integral of sqrt(2-2cos(x)) is -4cos(x/2)+C