How do you find the integral $\int x^{3} \cdot \sqrt{9 - x^{2}} d x$ $intx^3*sqrt(9-x^2)dx$ ?

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How do you find the integral $\int x^{3} \cdot \sqrt{9 - x^{2}} d x$ $intx^3*sqrt(9-x^2)dx$ ?

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Answer 1 · 2014-08-29T13:04:19

$= \frac{{(9 - x^{2})}^{\frac{5}{2}}}{5} - 3 {(9 - x^{2})}^{\frac{3}{2}} + c$ $=(9-x^2)^(5/2)/5-3(9-x^2)^(3/2)+c$ , where $c$ $c$ is a constant

Explanation :

$= \int x^{3} \cdot \sqrt{9 - x^{2}} d x$ $=intx^3*sqrt(9-x^2)dx$

Using Integration by Substitution,

let's assume $9 - x^{2} = t^{2}$ $9-x^2=t^2$ , then

$- 2 x d x = 2 t d t$ $-2xdx=2tdt$ $\Rightarrow x d x = - t d t$ $=>xdx=-tdt$

$= \int - (9 - t^{2}) t^{2} d t$ $=int-(9-t^2)t^2dt$

$= \int (t^{4} - 9 t^{2}) d t$ $=int(t^4-9t^2)dt$

$= \int t^{4} d t - 9 \int t^{2} d t$ $=intt^4dt-9intt^2dt$

$= \frac{t^{5}}{5} - 9 \frac{t^{3}}{3} + c$ $=t^5/5-9t^3/3+c$ , where $c$ $c$ is a constant

$= \frac{t^{5}}{5} - 3 t^{3} + c$ $=t^5/5-3t^3+c$ , where $c$ $c$ is a constant

Substituting $t$ $t$ back,

$= \frac{{(9 - x^{2})}^{\frac{5}{2}}}{5} - 3 {(9 - x^{2})}^{\frac{3}{2}} + c$ $=(9-x^2)^(5/2)/5-3(9-x^2)^(3/2)+c$ , where $c$ $c$ is a constant

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How do you find the integral $\int x^{3} \cdot \sqrt{9 - x^{2}} d x$ $intx^3*sqrt(9-x^2)dx$ ?

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How do you find the integral intx^3*sqrt(9-x^2)dx∫x3⋅√9−x2dxintx^3*sqrt(9-x^2)dx ?

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How do you find the integral $\int x^{3} \cdot \sqrt{9 - x^{2}} d x$ $intx^3*sqrt(9-x^2)dx$ ?