What is the integral of x^3sqrt(1-x^2) ?

The integral of x^3sqrt(1-x^2) is -1/15 (3x^2+2)(-x^2+1)^{3/2}+C

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integral of x^3sqrt(1-x^2)

\int x^{3} 1 - x^{2} d x

- \frac{1}{15} (3 x^{2} + 2) (- x^{2} + 1)^{\frac{3}{2}} + C

Solution steps

\int x^{3} 1 - x^{2} d x

Apply u-substitution

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

= \int - \frac{( - u + 1 ) 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 + 1) u d u

Expand

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

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

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

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

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

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

Substitute back

u = 1 - x^{2}

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

Simplify

- \frac{1}{2} (- \frac{2}{5} (1 - x^{2})^{\frac{5}{2}} + \frac{2}{3} (1 - x^{2})^{\frac{3}{2}}) : - \frac{1}{15} (3 x^{2} + 2) (- x^{2} + 1)^{\frac{3}{2}}

= - \frac{1}{15} (3 x^{2} + 2) (- x^{2} + 1)^{\frac{3}{2}}

Add a constant to the solution

= - \frac{1}{15} (3 x^{2} + 2) (- x^{2} + 1)^{\frac{3}{2}} + C

\frac{d}{d x} ((x + x)^{2})

y = \frac{1}{18 x ^{2}}, y = x, y = \frac{x}{6}

\int \frac{1}{- x ^{2} - a + b} d x

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

\frac{1}{3 s ^{2}} - \frac{5}{2 s}

What is the integral of x^3sqrt(1-x^2) ?
The integral of x^3sqrt(1-x^2) is -1/15 (3x^2+2)(-x^2+1)^{3/2}+C