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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

- a^{2} - y^{2} + C

Solution steps

\int \frac{y}{a ^{2} - 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 = a^{2} - y^{2}

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

Simplify

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

= - a^{2} - y^{2}

Add a constant to the solution

= - a^{2} - y^{2} + C

x \to 0 lim (x^{2} csc (x))

\frac{\partial}{\partial y} (3 x + 4 y)

x \to 0 lim (a^{x} + a^{- x})

\int (3 x + 4)^{6} d x

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