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Evaluate the definite integral

  • Thread starter sapiental
  • Start date
  • #1
118
0
the definite integral:

from a = (-pi/2) to (pi/2) f(((x^2)(sinx))/(1+x^6))dx

this is the way it seems most logic to me to set it up using substitution:

u = x
du = dx

from a = (-pi/2) to (pi/2) f(((u^2)(sinu))/(1+u^6))du

= (((-cos(u))(1/3u^3))/(u+1/7u^7))+C


I know how to evaluate it from here, I just need some feedback on my substitution setup.

Thanks in advance.
 

Answers and Replies

  • #2
2,063
2
sapiental said:
u = x
du = dx
That sort of substitution does nothing but change the letter denoting the variable.
 
  • #3
shmoe
Science Advisor
Homework Helper
1,992
1
Try sketching the graph. Notice anything?

courtrigrad said:
Take the square root of the numerator and denominator and use the substitution [tex] x^{3} = \tan \theta, x = \sqrt[3]{\tan \theta} [/tex] You end up getting [tex] \frac{1}{3} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \tan \theta [/tex]
I'm worried about what "Take the square root of the numerator and denominator" could possibly mean? I don't see how your substitution gives what you claim it gives either.
 
  • #4
1,235
1
yeah, I made a mistake. I took the integral of the square root of the function instead of the actual function.
 

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