What is the integral of y/(sqrt(16-y^2)) ?

The integral of y/(sqrt(16-y^2)) is -sqrt(16-y^2)+C

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integral of y/(sqrt(16-y^2))

\int \frac{y}{16 - y ^{2}} d y

- 16 - y^{2} + C

Solution steps

\int \frac{y}{16 - y ^{2}} d y

Apply u-substitution

= \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

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

Substitute back

u = 16 - y^{2}

= - \frac{1}{2} \cdot 2 (16 - y^{2})^{\frac{1}{2}}

Simplify

- \frac{1}{2} \cdot 2 (16 - y^{2})^{\frac{1}{2}} : - 16 - y^{2}

= - 16 - y^{2}

Add a constant to the solution

= - 16 - y^{2} + C

What is the integral of y/(sqrt(16-y^2)) ?
The integral of y/(sqrt(16-y^2)) is -sqrt(16-y^2)+C