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Evaluate the definite integral

  • Thread starter sapiental
  • Start date
  • #1
118
0
hi,

I've been having difficulty with this integral for some time now and any help would be gratly appreciated.

[tex]\int\frac{x^2 \sin x}{1+x^6}dx[/tex]

this is a definite integral from -pi/2 to pi/2

The sinx has been giving me problems because if I set u = to any part of the equation I can't write sin(u)

for example

u = x^2 du/2 = xdx
u = 1+x^6 du/6 = x^5dx
u = 1+x^3 du/2 = x^2dx

in all these cases I still get stuck with the sinx..

hints on how to approach this equation would be ideal becasue I need to learn how to do this myself.

Thank you!!
 

Answers and Replies

  • #2
StatusX
Homework Helper
2,564
1
You won't be able to get the indefinite integral for that thing. But fortunately, you can use the fact the function is odd and the integration region is even.
 
  • #3
118
0
hey thanks for the help everyone.

StatusX, what you're saying is that I can just write the final result as

[tex]\int \frac{x^2 \sin x}{1+x^6}dx = 0 [/tex]

and the function is odd because of sin(x) right?


thanks
 
Last edited:
  • #4
StatusX
Homework Helper
2,564
1
Yea, because it is an even function times an odd function.
 

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