What is the integral of usqrt(1-u^2) ?

The integral of usqrt(1-u^2) is -1/3 (1-u^2)^{3/2}+C

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integral of usqrt(1-u^2)

\int u 1 - u^{2} d u

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

Solution steps

\int u 1 - u^{2} d u

Apply u-substitution

: \int - v^{2} d v

= \int - v^{2} d v

Take the constant out:

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

= - \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

= - \frac{v ^{3}}{3}

Substitute back

v = 1 - u^{2}

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

Simplify

- \frac{( 1 - u ^{2} ) ^{3}}{3} : - \frac{1}{3} (1 - u^{2})^{\frac{3}{2}}

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

Add a constant to the solution

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

\int x \cdot e^{- 2 x} d x

\frac{d}{d x} (\frac{1}{10 x})

14 - x^{2}, x^{2} - 4

\frac{d y}{d x} + 4 x y^{4} = 0

\frac{( x - 5 )}{x + 5}

What is the integral of usqrt(1-u^2) ?
The integral of usqrt(1-u^2) is -1/3 (1-u^2)^{3/2}+C