What is the integral of x(1-x)^{1/2} ?

The integral of x(1-x)^{1/2} is -2(-1/5 (1-x)^{5/2}+1/3 (1-x)^{3/2})+C

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integral of x(1-x)^{1/2}

integral

x (1 - x)^{\frac{1}{2}}

- 2 (- \frac{1}{5} (1 - x)^{\frac{5}{2}} + \frac{1}{3} (1 - x)^{\frac{3}{2}}) + C

Solution steps

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

Apply u-substitution

: \int - 2 (- u^{2} + 1) u^{2} d u

= \int - 2 (- u^{2} + 1) u^{2} d u

Take the constant out:

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

= - 2 \cdot \int (- u^{2} + 1) u^{2} d u

Expand

(- u^{2} + 1) u^{2} : - u^{4} + u^{2}

= - 2 \cdot \int - u^{4} + u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 2 (- \int u^{4} d u + \int u^{2} d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int u^{2} d u = \frac{u ^{3}}{3}

= - 2 (- \frac{u ^{5}}{5} + \frac{u ^{3}}{3})

Substitute back

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

= - 2 - \frac{( ( 1 - x ) ^{\frac{1}{2}} ) ^{5}}{5} + \frac{( ( 1 - x ) ^{\frac{1}{2}} ) ^{3}}{3}

Simplify

- 2 - \frac{( ( 1 - x ) ^{\frac{1}{2}} ) ^{5}}{5} + \frac{( ( 1 - x ) ^{\frac{1}{2}} ) ^{3}}{3} : - 2 (- \frac{1}{5} (1 - x)^{\frac{5}{2}} + \frac{1}{3} (1 - x)^{\frac{3}{2}})

= - 2 (- \frac{1}{5} (1 - x)^{\frac{5}{2}} + \frac{1}{3} (1 - x)^{\frac{3}{2}})

Add a constant to the solution

= - 2 (- \frac{1}{5} (1 - x)^{\frac{5}{2}} + \frac{1}{3} (1 - x)^{\frac{3}{2}}) + C

\frac{x}{( 1 - x ) ^{2}}

\frac{( 1 - 3 x ) ^{2}}{x ^{3}}

\frac{x ^{3}}{( 1 + x ^{2} ) ^{2}}

sin^{2} (x) + cos^{2} (x)

\frac{ln ( x )}{( x + 1 ) ^{2}}

What is the integral of x(1-x)^{1/2} ?
The integral of x(1-x)^{1/2} is -2(-1/5 (1-x)^{5/2}+1/3 (1-x)^{3/2})+C