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anonymous

  • one year ago

solve this integral :/ not sure how to do this!

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  1. anonymous
    • one year ago
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    \[\int\limits_{} \sin^5 t \cos^4 t dt\]

  2. Zale101
    • one year ago
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    |dw:1442984817219:dw|

  3. Astrophysics
    • one year ago
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    Some general rules with these trig functions |dw:1442984972414:dw|

  4. Zale101
    • one year ago
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    Any ideas of what to do next?

  5. Zale101
    • one year ago
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    That chart is very helpful! @astrophysics

  6. Astrophysics
    • one year ago
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    Yeah, it's pretty awesome :P

  7. Astrophysics
    • one year ago
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    http://puu.sh/kl22A/732d78a86f.png if anyone wants the link XD

  8. anonymous
    • one year ago
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    sorry!! i am back!! :) i am reading the chart right now :) one sec!! hopefully i can have a clue what happens next:)

  9. anonymous
    • one year ago
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    okay, i see now what you did @Zale101 so i strip one sin away and get|dw:1442985650253:dw|

  10. anonymous
    • one year ago
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    not sure what to do next though :/

  11. Astrophysics
    • one year ago
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    |dw:1442985664022:dw|

  12. anonymous
    • one year ago
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    ohh so i would get this? |dw:1442985687508:dw|

  13. Zale101
    • one year ago
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    |dw:1442985746191:dw|

  14. anonymous
    • one year ago
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    okay!! i forget though, from here, do i sub something in? :/

  15. Zale101
    • one year ago
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    U-Sub! :)

  16. anonymous
    • one year ago
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    so we get integral sin t (sin^2 t)^2 u^4 dt ?

  17. Zale101
    • one year ago
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    Nope. First of all, you cant have u while the rest of the terms are in t. Second, you have to use trig identities for (sin^2t)^2

  18. anonymous
    • one year ago
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    ohhh how do i do that then?

  19. Zale101
    • one year ago
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    \((sin^2t)^2=(1-cos^2t)^2\)

  20. anonymous
    • one year ago
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    ohh the square thing?

  21. Zale101
    • one year ago
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    \(sin^4t=(sin^2t)^2=(1−cos^2t)^2\)

  22. anonymous
    • one year ago
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    confused which trig functions to insert in now?

  23. Zale101
    • one year ago
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    Yes. If you wan to replace sin^2 with the trig identities 1-cos^2t. But sin is squared, so the identity has to be squared as well.

  24. anonymous
    • one year ago
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    ohh but it is fine leaving like sin? what happens next?

  25. Zale101
    • one year ago
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    Next, you do u-sub. But first show me your integral now after using trig identities.

  26. anonymous
    • one year ago
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    \[\int\limits_{} (1-\cos^2 t)((1-\cos^2)^2 t)^2 cost^4 t dt\] this?

  27. Zale101
    • one year ago
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    \[\int\limits_{}^{}\sin(t) (\sin^2(t))^2\cos^4(t)dt\] \[\int\limits_{}^{}\sin(t) (1-cos^2(t))^2\cos^4(t)dt\]

  28. anonymous
    • one year ago
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    ohh okay oopsies :P what happens now? :O

  29. Zale101
    • one year ago
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    What you did was wrong because you can replace sint with 1-cos^2 trig identities say \(sin^2t=1-cos^2\)

  30. anonymous
    • one year ago
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    ohhhh:O

  31. Zale101
    • one year ago
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    \(sin^2t+cos^2t=1\)

  32. Zale101
    • one year ago
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    What is your next step now?

  33. anonymous
    • one year ago
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    i integrate? :/

  34. Zale101
    • one year ago
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    How?

  35. anonymous
    • one year ago
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    i am not sure :(

  36. anonymous
    • one year ago
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    i am sorry :( i am very bad at calc:(

  37. Zale101
    • one year ago
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    We need u-sub here. What would be you u in this case?

  38. Zale101
    • one year ago
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    your*

  39. anonymous
    • one year ago
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    u=sin(t) ?

  40. Zale101
    • one year ago
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    Would u=sint work?

  41. anonymous
    • one year ago
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    i think so :/

  42. Zale101
    • one year ago
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    I will try u=sint and see if it works.

  43. anonymous
    • one year ago
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    ok!! sorry, i get very confused with these concepts :/

  44. Zale101
    • one year ago
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    |dw:1442986993486:dw|

  45. anonymous
    • one year ago
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    ohhh okay :/ what should we do then?

  46. Zale101
    • one year ago
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    try u=cos(t)

  47. Astrophysics
    • one year ago
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    A saying I remember, get the odd man out! You want the odd man out!

  48. anonymous
    • one year ago
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    okay!! where do we sub that into ? :/

  49. Zale101
    • one year ago
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    With integration, here is what you should do. Every time you make an educated guess on what might the u be, do the integration and see if it works.If not, then try something else. This is what i did with integral and it was helpful.

  50. Zale101
    • one year ago
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    integral problems*

  51. anonymous
    • one year ago
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    ohhh okie :) so we should use u=cos(t) and so we do this? \[\int\limits_{}sint(1-u)^2 u^4 dt \]

  52. Zale101
    • one year ago
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    You cant place in the u when you have not finished u-subbing

  53. anonymous
    • one year ago
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    ohh no! I'm sorry, i don't know what to do here :(

  54. Astrophysics
    • one year ago
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    When you do a u sub, you must also get the derivative, that's what will cancel out the sint \[u= cost \implies du= - sint dt\]

  55. Astrophysics
    • one year ago
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    You don't want t and u's, just one variable

  56. anonymous
    • one year ago
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    okay! what would it look like then?

  57. Zale101
    • one year ago
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    Do u=cost and du=-sint dt and tell me what you will get

  58. anonymous
    • one year ago
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    i get this? is it correct? integral with plug in u and du udu and i sub in and replace and get this? -1/5cos^5t+2/7cost^7t-1/9cos^9t+C?

  59. anonymous
    • one year ago
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    thank you so much @Zale101 @Astrophysics i will try to figure this one out some more tomorrow! i need to sleep now because it is 2am here and i have to be up in four and a half hrs!! thanks so much for your help!!! :D sorry it was hard for me to understand!! have a nice night!!

  60. Zale101
    • one year ago
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    Alright, good luck and take care!

  61. Astrophysics
    • one year ago
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    Good night! :)

  62. Zale101
    • one year ago
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    Just a type error i made "What you did was wrong because you \(\color{red}{can't}\) replace sint with 1-cos^2"

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