Trigonometric Integrals

  • Thread starter bgut06
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  • #1
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Homework Statement



integral 31(cos^2x)(sin(2x)dx

Homework Equations





The Attempt at a Solution


I am so lost on this problem... Any suggestions would be great
 

Answers and Replies

  • #2
phyzguy
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Try looking up the double angle identities and use them to re-write sin(2x). Then try integration by substitution.
 
  • #3
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So if the double angle for sin(2x) is 2sinxcosx... the problem would be rewritten as integral 31cos^2x(2sinxcosx)dx? Then what??? Sorry im still confused

Thanks for the help!
 
  • #4
phyzguy
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Now collect the cos(x) terms together. Then what do you have?
 
  • #5
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So it would become integral 31(2sinxcos^3)... Then make u=sinx du=cosxdx... Which would lead to 31 integral 2sinxcos^2xcosxdx... Then 31 integral 2sinx(1-sin^2x)du... 31 integral 2u(1-u^2)du... Eventually leading to 31u^2-31/2u^4... And then 31sin^2x-31/2sin^4x.. How does that look??
 
  • #6
phyzguy
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It's much simpler if you try u=cos(x), but I think your solution is correct.
 
Last edited:
  • #7
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Awesome! Thank you so much. I have another question if you don't mind.. . For Integral 3x(cos(2x))^2dx could I use u substitution with the 3x
 

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