Trigonometric integral

  • Thread starter Stevecgz
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  • #1
Stevecgz
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Problem:
[tex]\int sin^6 x dx[/tex]
Progress so far:
[tex]\int (sin^2 x)^3 dx[/tex]
[tex]\frac{1}{8} \int (1-cos2x)^3 dx [/tex]
[tex]\frac 1 8 \int (1 - 3cos2x + 3cos^22x - cos^32x) dx[/tex]

Any help is appreciated.

I can see using a half angle identity for cos^2(2x), but what do I do with the cos^3(2x)?


Steve
 
Last edited:

Answers and Replies

  • #2
whozum
2,221
1
Try looking up some sine integral reduction formulas on google. They take care of integrals involving powers of sine pretty nicely.
 
  • #3
Stevecgz
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whozum said:
Try looking up some sine integral reduction formulas on google. They take care of integrals involving powers of sine pretty nicely.

I've found one in my text. Would I simply continue using the reduction formula until I get to sin^0(x)?

Steve
 
  • #4
whozum
2,221
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Yes that's pretty much how we did it.
 
  • #5
Stevecgz
68
0
Thanks whozum.

Steve
 

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