What is the integral of (2r)/(sqrt(4-r^2)) ?

The integral of (2r)/(sqrt(4-r^2)) is -2sqrt(4-r^2)+C

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integral of (2r)/(sqrt(4-r^2))

\int \frac{2 r}{4 - r ^{2}} d r

- 2 4 - r^{2} + C

Solution steps

\int \frac{2 r}{4 - r ^{2}} d r

Take the constant out:

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

= 2 \cdot \int \frac{r}{4 - r ^{2}} d r

Apply u-substitution

: \int - \frac{1}{2 u} d u

= 2 \cdot \int - \frac{1}{2 u} d u

Take the constant out:

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

= 2 (- \frac{1}{2} \cdot \int \frac{1}{u} d u)

Apply the Power Rule

: 2 u^{\frac{1}{2}}

= 2 (- \frac{1}{2} \cdot 2 u^{\frac{1}{2}})

Substitute back

u = 4 - r^{2}

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

Simplify

2 (- \frac{1}{2} \cdot 2 (4 - r^{2})^{\frac{1}{2}}) : - 2 4 - r^{2}

= - 2 4 - r^{2}

Add a constant to the solution

= - 2 4 - r^{2} + C

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