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

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

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

\int x^{3} \cdot 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

y = x^{2} (x - 11), [- 1, 12]

n = 1 \sum \infty \frac{1}{2 n + 7}

\frac{1}{( s ^{2} + 4 ) ( s ^{2} + 2 s + 2 )}

2 t \cdot sin (t)

t y^{^{'}} - y = \frac{- t ^{3}}{y ^{2}}

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