What is the integral of 5x^3sqrt(1-x^2) ?

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

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integral of 5x^3sqrt(1-x^2)

\int 5 x^{3} 1 - x^{2} d x

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

Solution steps

\int 5 x^{3} 1 - x^{2} d x

Take the constant out:

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

= 5 \cdot \int x^{3} 1 - x^{2} d x

Apply u-substitution

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

= 5 \cdot \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

= 5 (- \frac{1}{2} \cdot \int (- u + 1) u d u)

Expand

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

= 5 (- \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

= 5 (- \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}}

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

Substitute back

u = 1 - x^{2}

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

Add a constant to the solution

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

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\frac{\partial}{\partial x} (\frac{( x ( y + 3 ) )}{x ^{2} + y})

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