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How do you find the integral #intx^3*sqrt(9-x^2)dx# ?

Answer 1

Emma Aldridge

#=(9-x^2)^(5/2)/5-3(9-x^2)^(3/2)+c#, where #c# is a constant

Explanation :

#=intx^3*sqrt(9-x^2)dx#

Using Integration by Substitution,

let's assume #9-x^2=t^2#, then

#-2xdx=2tdt# #=>xdx=-tdt#

#=int-(9-t^2)t^2dt#

#=int(t^4-9t^2)dt#

#=intt^4dt-9intt^2dt#

#=t^5/5-9t^3/3+c#, where #c# is a constant

#=t^5/5-3t^3+c#, where #c# is a constant

Substituting #t# back,

#=(9-x^2)^(5/2)/5-3(9-x^2)^(3/2)+c#, where #c# is a constant

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Answer 2

Christopher Agresta

To find the integral of ( \int x^3 \sqrt{9 - x^2} , dx ), you can use the trigonometric substitution method. Let ( x = 3 \sin(\theta) ), then ( dx = 3 \cos(\theta) , d\theta ). Substitute these into the integral and simplify. You'll get ( \int 27 \sin^3(\theta) \cos^2(\theta) , d\theta ). Use the identity ( \sin^2(\theta) = 1 - \cos^2(\theta) ) to express everything in terms of ( \cos(\theta) ), then integrate. The final answer involves trigonometric functions.

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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.