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Dido525

  • one year ago

Evaluate the intergal

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  1. Dido525
    • one year ago
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    \[\int\limits_{}^{}3\sin^2(t)\cos^4(t)dt\]

  2. Dido525
    • one year ago
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    My work: \[3\int\limits_{}^{}\sin^2(x)\cos^4(t)dt\] \[3\int\limits_{}^{}\sin^2(t)\cos^2(t)\cos^2(t)dt\] \[3\int\limits_{}^{}\frac{ 1 }{ 2 }(1-\cos(2t))(\frac{ 1 }{ 2 }(1+\cos(2t)))^2dt\] \[\frac{ 3 }{ 8 }\int\limits_{}^{}(1-\cos(2t))((1+\cos(2t)))^2dx\] \[\frac{ -3 }{ 8 }\int\limits_{}^{}((\cos^3(2t)+\cos^2(2t)+\cos(2t)+1) dt\] \[\frac{ -3 }{ 8 }\int\limits_{}^{}((cos^2(2t))(cos(2t))+\cos^2(2t)+\cos(2t)+1))dt\] \[\frac{ -3 }{ 8 }\int\limits_{}^{}(\cos^3(2t)+\cos^2(2t)+\cos(2t)+1)dt\]

  3. Dido525
    • one year ago
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    I don't know what to do after :( .

  4. Dido525
    • one year ago
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    The 4th step should be a dt.

  5. satellite73
    • one year ago
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    look in the back of your book at the "reduction" formula i think you can make a u sub earlier on though. maybe i am wrong

  6. Dido525
    • one year ago
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    Is there no other way?

  7. Dido525
    • one year ago
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    I mean I did get a little further.

  8. Dido525
    • one year ago
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    \[\frac{ -3 }{ 8 }\int\limits\limits_{}^{}(\cos^2(2t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt\]

  9. satellite73
    • one year ago
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    i would start with \[\int\cos^4(x)dx-\int\cos^6(x)dx\] and then look in the back of the book

  10. Dido525
    • one year ago
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    \[\frac{ -3 }{ 8 }\int\limits\limits\limits_{}^{}((1-\sin^2(t)\cos(2t)+\cos^2(2t)+\cos(2t)+1)dt\]

  11. Dido525
    • one year ago
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    Where did that come from?

  12. satellite73
    • one year ago
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    there are fomulas that i cannot remember for integrating \(\sin^n(x)\) and \(\cos^n(x)\) and i garantee you they are on the back cover of your text

  13. Dido525
    • one year ago
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    Yeah, I know. I turn them into the half angles, which I did.

  14. Dido525
    • one year ago
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    That stupid cos^2(2t) is getting in the way.

  15. satellite73
    • one year ago
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    \(\sin^2(x)=1-\cos^2(x)\) then multiply out it is easier than reinventing the wheel you are trying to derive the formula, (which is admirable) but it is easier to look it up

  16. Dido525
    • one year ago
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    Using another half angle won't help because they are multiplied together which just makes it squared afain,

  17. Dido525
    • one year ago
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    again*

  18. satellite73
    • one year ago
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    why i find this topic rather dull. almost everything you need is printed on the back cover of your text

  19. Dido525
    • one year ago
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    I know. I have to do it though.

  20. satellite73
    • one year ago
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    then i really recommend looking up the formula for \(\int\cos^n(x)dx\)

  21. Dido525
    • one year ago
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    So I get:\[\frac{ -3 }{ 8 }\int\limits_{}^{}(1-\sin^2(2t))\cos(2t)+(1-\sin^2(2t)+\cos(2t)+1)dt\]

  22. Dido525
    • one year ago
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    Hmm good point.

  23. Dido525
    • one year ago
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    Sec.

  24. Dido525
    • one year ago
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    Can I ask a dumb question?

  25. Dido525
    • one year ago
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    We have that constant out in front. If we break up the intergal into seperate sums do I distribute that constant to all the intergals?

  26. satellite73
    • one year ago
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    sure so i can seem dumber if i cannot answer it

  27. satellite73
    • one year ago
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    yes (whew)

  28. satellite73
    • one year ago
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    ignore the annoying 3, put it in at the end

  29. Dido525
    • one year ago
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    Hehe. Thanks. This should be easy now.

  30. satellite73
    • one year ago
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    here it is \[\int\cos^n(x)dx = \frac{1}{n}\cos(x)\sin(x)+\frac{n-1}{n}\int \cos^{n-2}(x)dx\] use with \(n=4\) and again with \(n=6\)

  31. Dido525
    • one year ago
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    Ohh that's MASSIVELY convenient.

  32. satellite73
    • one year ago
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    it is going to be a pain, but much less of a pain than what you were doing

  33. Dido525
    • one year ago
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    Why 4 and 6?

  34. Dido525
    • one year ago
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    I have 3 and 2.

  35. satellite73
    • one year ago
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    \[\sin^2(x)\cos^4(x)=(1-\cos^2(x))\cos^4(x)=\cos^4(x)-\cos^6(x)\]

  36. satellite73
    • one year ago
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    so 4 and 6

  37. Dido525
    • one year ago
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    Right.

  38. Dido525
    • one year ago
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    Thanks!

  39. satellite73
    • one year ago
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    have fun!

  40. satellite73
    • one year ago
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    yw

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