$\int x^{3} \sqrt{1 - x^{2}} d x$ $\intx^3\sqrt(1-x^2)dx$ ?

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$\int x^{3} \sqrt{1 - x^{2}} d x$ $\intx^3\sqrt(1-x^2)dx$ ?

I tried to get the answer found on Symbolab, but instead I got:
$- \frac{1}{3} {(1 - x^{2})}^{\frac{3}{2}} + \frac{1}{5} {(1 - x^{2})}^{\frac{5}{2}} + C$ $-1/3(1-x^2)^(3/2)+1/5(1-x^2)^(5/2)+C$

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Answer 1 · 2018-04-13T04:51:25

Your answer is excellent and correct!

Explanation:

You're going to have to use trig substitution for this question.
Let $x = sin θ$ $x= sintheta$ . Then $d x = cos θ d θ$ $dx =costheta d theta$ .

$I = \int {sin}^{3} θ \sqrt{1 - {sin}^{2} θ} cos θ d θ$ $I = int sin^3theta sqrt(1 - sin^2theta) costheta d theta$

$I = \int {sin}^{3} θ \sqrt{{cos}^{2} θ} cos θ d θ$ $I = int sin^3theta sqrt(cos^2theta)costheta d theta$

$I = \int {sin}^{3} θ {cos}^{2} θ d θ$ $I = int sin^3theta cos^2thetad theta$

$I = \int sin θ (1 - {cos}^{2} θ) {cos}^{2} θ d θ$ $I = int sintheta(1 -cos^2theta)cos^2thetad theta$

$I = \int (sin θ - sin θ {cos}^{2} θ) {cos}^{2} θ d θ$ $I = int (sin theta - sinthetacos^2theta)cos^2thetad theta$

$I = \int sin θ {cos}^{2} θ - sin θ {cos}^{4} θ d θ$ $I = int sinthetacos^2theta - sinthetacos^4thetad theta$

$I = \int sin θ {cos}^{2} θ d θ - \int sin θ {cos}^{4} θ d θ$ $I= int sin thetacos^2theta d theta - int sinthetacos^4theta d theta$

Let $u = cos θ$ $u = costheta$ . You should be left with $d u = - sin θ d θ$ $du = -sintheta d theta$ and $d θ = \frac{d u}{- sin θ}$ $d theta = (du)/(-sintheta)$ .

$I = - \int u^{2} d u + \int u^{4} d u$ $I = -int u^2 du +int u^4 du$

$I = - \frac{1}{3} u^{3} + \frac{1}{5} u^{5} + C$ $I = -1/3u^3 + 1/5u^5 + C$

$I = - \frac{1}{3} {cos}^{3} θ + \frac{1}{5} {cos}^{5} θ + C$ $I = -1/3cos^3theta + 1/5cos^5theta + C$

Recall from our initial substitution that $x = sin θ$ $x= sintheta$ , and since ${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x+ sin^2x = 1$ , we can see that $cos θ = \sqrt{1 - x^{2}}$ $costheta = sqrt(1 - x^2)$ .

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

I checked and our answer is the same as the one shown on symbolab, except ours is simplified a little further.

Hopefully this helps!

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$\int x^{3} \sqrt{1 - x^{2}} d x$ $\intx^3\sqrt(1-x^2)dx$ ?

I tried to get the answer found on Symbolab, but instead I got:
$- \frac{1}{3} {(1 - x^{2})}^{\frac{3}{2}} + \frac{1}{5} {(1 - x^{2})}^{\frac{5}{2}} + C$ $-1/3(1-x^2)^(3/2)+1/5(1-x^2)^(5/2)+C$

1 Answer

Explanation:

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∫x3√1−x2dx\intx^3\sqrt(1-x^2)dx?

I tried to get the answer found on Symbolab, but instead I got: −13(1−x2)32+15(1−x2)52+C-1/3(1-x^2)^(3/2)+1/5(1-x^2)^(5/2)+C

1 Answer

Explanation:

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Impact of this question

$\int x^{3} \sqrt{1 - x^{2}} d x$ $\intx^3\sqrt(1-x^2)dx$ ?

I tried to get the answer found on Symbolab, but instead I got:
$- \frac{1}{3} {(1 - x^{2})}^{\frac{3}{2}} + \frac{1}{5} {(1 - x^{2})}^{\frac{5}{2}} + C$ $-1/3(1-x^2)^(3/2)+1/5(1-x^2)^(5/2)+C$