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

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

Popular >

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

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

- 4 - y^{2} + C

Solution steps

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

Apply u-substitution

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

= \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

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

Apply the Power Rule

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

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

Substitute back

u = 4 - y^{2}

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

Simplify

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

= - 4 - y^{2}

Add a constant to the solution

= - 4 - y^{2} + C

\frac{\partial}{\partial z} (5 x^{2} - 3 x y + x yz)

\frac{\partial}{\partial t} (ln (3 x + 3 c t))

x \to 0 lim (\frac{1}{x} - \frac{1}{e ^{x}})

\frac{1}{( s ^{2} + 1 ) ( s ^{2} + 16 )}

x \to 2 lim (x^{2} - 4 x + 5)

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