What is the integral of rsqrt(36-r^2) ?

The integral of rsqrt(36-r^2) is -1/3 (36-r^2)^{3/2}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of rsqrt(36-r^2)

\int r 36 - r^{2} d r

- \frac{1}{3} (36 - r^{2})^{\frac{3}{2}} + C

Solution steps

\int r 36 - r^{2} d r

Apply u-substitution

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

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

Apply the Power Rule

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

= - \frac{1}{2} \cdot \frac{2}{3} u^{\frac{3}{2}}

Substitute back

u = 36 - r^{2}

= - \frac{1}{2} \cdot \frac{2}{3} (36 - r^{2})^{\frac{3}{2}}

Simplify

- \frac{1}{2} \cdot \frac{2}{3} (36 - r^{2})^{\frac{3}{2}} : - \frac{1}{3} (36 - r^{2})^{\frac{3}{2}}

= - \frac{1}{3} (36 - r^{2})^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{3} (36 - r^{2})^{\frac{3}{2}} + C

\int 2 - y d y

\int \frac{x + 3}{x ^{2} + 2 x - 5} d x

\int \frac{x ^{3}}{12} d x

\int \frac{( 3 + 2 x )}{4 e ^{4 x}} d x

\int \frac{( x + 1 )}{4 x ^{2} - 4 x + 1} d x

What is the integral of rsqrt(36-r^2) ?
The integral of rsqrt(36-r^2) is -1/3 (36-r^2)^{3/2}+C