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How do you find the integral #intx^3/(sqrt(16-x^2))dx# ?

Answer 1

Christopher Agresta

#=(16-x^2)^(3/2)/3-16sqrt(16-x^2)+c#, where #c# is a constant

Solution

#=intx^3/sqrt(16-x^2)dx=int(x^2*x)/sqrt(16-x^2)dx#

Using Integration by Substitution

let's assume #(16-x^2)=t^2#, #=>-2xdx=2tdt#

#xdx=-tdt#

#=int(-(16-t^2)t)/tdt#

#=int(t^2-16)dt#

#=intt^2dt-16intdt#

#=t^3/3-16t+c#, where #c# is a constant

#=(16-x^2)^(3/2)/3-16sqrt(16-x^2)+c#, where #c# is a constant

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Answer 2

Paisley Johnson

To find the integral of ( \int \frac{x^3}{\sqrt{16 - x^2}} , dx ), you can use the trigonometric substitution method. Let ( x = 4\sin(\theta) ), then ( dx = 4\cos(\theta) , d\theta ). Substituting these into the integral and simplifying, you'll eventually obtain:

[ \int \frac{x^3}{\sqrt{16 - x^2}} , dx = -4\left(\frac{1}{2}\right) \cos^2(\theta) , d\theta ]

After integrating with respect to ( \theta ), you'll have:

[ -2\theta - \frac{1}{2}\sin(2\theta) + C ]

Finally, substitute ( \theta ) back in terms of ( x ) using ( x = 4\sin(\theta) ) and simplify the expression to get the result.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.