Integration with Trigonometric Substitution

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[SOLVED] Integration with Trigonometric Substitution

Homework Statement



Given integral (I):
I[(x)sqrt(9-x^2)dx]

by words:
Integral of "X" times square root of "9-X(squared)

Use proper trigonometric substitution to solve this problem.

Homework Equations





The Attempt at a Solution

 

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You don't even need Trig substitution.

\int x\sqrt{9-x^2}dx

u=9-x^2
du=-2xdx \rightarrow xdx=-\frac 1 2 du
 
I know that I don't need that.
But the problem is, I have to use it.
The exercise require it.
 
Well you posted the solution to it? I don't know what else to tell you. Just analyze what they did. Work it yourself a couple times if you have to.
 
You forgot a term when you first did the substitution.

x = 3sin(u)
dx = 3cos(u)du

(9 - x^2)^(1/2) = 3cos(u)

So the integral becomes 27sin(u)cos^2(u)du
 
\int x^2\sqrt{9-x^2}dx
apropo
u=x^2\sqrt{9-x^2}dx
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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