What is the integral of sqrt(8-8cos(x)) ?

The integral of sqrt(8-8cos(x)) is -8cos(x/2)+C

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integral of sqrt(8-8cos(x))

\int 8 - 8 cos (x) d x

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

Solution steps

\int 8 - 8 cos (x) d x

Apply u-substitution

: \int \frac{8 u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

= \int \frac{8 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

= 8 \cdot \int \frac{u}{( 1 + u ^{2} ) 1 + u ^{2}} d u

Apply u-substitution

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

= 8 \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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{v v} d v

Simplify

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

= 8 \cdot \frac{1}{2} \cdot \int \frac{1}{v ^{\frac{3}{2}}} d v

Apply the Power Rule

: - \frac{2}{v}

= 8 \cdot \frac{1}{2} (- \frac{2}{v})

Substitute back

= 8 \cdot \frac{1}{2} - \frac{2}{1 + tan ^{2} ( \frac{x}{2} )}

Simplify

8 \cdot \frac{1}{2} - \frac{2}{1 + tan ^{2} ( \frac{x}{2} )} : - 8 cos (\frac{x}{2})

= - 8 cos (\frac{x}{2})

Add a constant to the solution

= - 8 cos (\frac{x}{2}) + C

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What is the integral of sqrt(8-8cos(x)) ?
The integral of sqrt(8-8cos(x)) is -8cos(x/2)+C