What is the integral of sqrt(1-x^2)x ?

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

integral of sqrt(1-x^2)x

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

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

Solution steps

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

Apply u-substitution

= \int - u^{2} d u

Take the constant out:

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

= - \int u^{2} d u

Apply the Power Rule

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

Substitute back

u = 1 - x^{2}

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

Simplify

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

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

Add a constant to the solution

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

a re a 20 x + x^{2} - x^{3}, y = 0, [- 4, 5]

a re a x^{2} + 4 x - 5, - x^{2} + 4 x + 3

x \to 1 lim (4 x^{2})

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

t an g e n t \frac{x ^{4} - x + 1}{x ^{4} + x + 1}

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