Integrate w/negative exponent

  • Thread starter anderma8
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  • #1
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I need a little 'suggestion' as to how to integrate cos^6x sin^-3x dx.

I rewrite to cos^6x/sin^3x dx and let u = sinx but when I'm trying to rewrite integral, what should I do with the ^6?

Thanks!
 

Answers and Replies

  • #2
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oh wait, re-reading the problem and I think what I'm interpreting as a negative is the dot in the i in sin.... I'm going to work the problem as cos^6x sin^3x dx and I know how to work this... sorry about that!
 
  • #3
HallsofIvy
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For future reference, if you did have an odd power of sine in the denominator, multiply both numerator and denominator by sine:
[tex]\int \frac{cos^6 t}{sin^3 t}dt= \frac{cos ^6 t sin t}{sin^4 t}dt[/tex]
[tex]= \int \frac{cos^6 t}{(1- cos^2 t)^2}(sin t dt)[/tex]
so that the substitution x= cos t gives
[tex]-\int\frac{x^6}{(1-x^2)^2}dx[/tex]
 

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