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anonymous

  • 5 years ago

how do you find the integral of sin^3x?

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  1. anonymous
    • 5 years ago
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    -3cos(3x)

  2. anonymous
    • 5 years ago
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    the answer is -cosx + (cos^3x/3) + c how do you get there?

  3. anonymous
    • 5 years ago
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    oops didnt see the ^

  4. anonymous
    • 5 years ago
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    so then im guessing what its SUPPOSED to be is sin(x)^3?

  5. anonymous
    • 5 years ago
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    yes that would be another way of writing the question

  6. anonymous
    • 5 years ago
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    \[\int\limits_{}^{} \sin^3xdx=\int\limits_{}^{}(1-\cos^2x)(sinx)dx\]

  7. anonymous
    • 5 years ago
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    So then use then let u=cosx, so then du=-sinxdx.

  8. anonymous
    • 5 years ago
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    What I do not understand is why u=cosx?! How did you know to choose that?

  9. anonymous
    • 5 years ago
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    So then you have, \[-\int\limits_{}^{}1-u^2du\]

  10. anonymous
    • 5 years ago
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    Essentially you want to get it to a point where you can work with it.

  11. anonymous
    • 5 years ago
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    Since I want to get rid of the sin stuff, so to speak, I substitute cosx to get sinx grouped with dx

  12. anonymous
    • 5 years ago
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    It's harder to explain via computer though. :/

  13. anonymous
    • 5 years ago
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    I'm assuming you can take the integral of the one I typed, correct?

  14. anonymous
    • 5 years ago
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    yes, but where did the 1 from 1- u^2 come from? Sorry, this is hard for me to grasp

  15. anonymous
    • 5 years ago
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    You could have also done it this way: \[\int\limits_{}^{} \sin^3xdx=\int\limits_{}^{}(1-\sin^2x)(cosx)dx\]

  16. anonymous
    • 5 years ago
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    AH! Forgot something.

  17. anonymous
    • 5 years ago
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    You could have substituted 1-cosx^2 for u as well

  18. anonymous
    • 5 years ago
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    \[\int\limits_{}\sin^3(x)dx=\int\limits_{}\sin^2(x)\sin(x)dx\] \[\int\limits_{}(1-\cos^2(x))\sin(x)dx\] \[\int\limits_{}\sin(x)dx-\int\limits{}\cos^2(x)dx\] use u substitution\[u=\cos(x), du=-\sin(x)dx, -du=\sin(x)dx\] \[-\cos(x)+\int\limits_{}u^2du=-\cos(x)+1/3(\cos^3(x))+c\] \[\cos(x)[1/3(\cos^2(x))-1] +c\]

  19. anonymous
    • 5 years ago
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    What he said. ^ =D

  20. anonymous
    • 5 years ago
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    sorry mistake \[\int\limits_{}\sin(x)dx=\int\limits_{}\cos^2(x)\sin(x)dx\] then u substitution

  21. anonymous
    • 5 years ago
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    cos(x)[1/3(cos2(x))−1]+c I don't understand this step Where does the -1 come from? this is before the final answer right?

  22. anonymous
    • 5 years ago
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    no that is the final answered I factored out a cos(x) because both terms in the final answer have cos(x)

  23. anonymous
    • 5 years ago
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    if du = -sinxdx and u= cosx Then you make it -cosx - integral of u^2 thennn you get -cosx -cos3x/3? how did you get -1 to make it -cosx+cos3x.....

  24. anonymous
    • 5 years ago
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    if u = cosx then where did the cosxsinx go? I am so confusedd

  25. anonymous
    • 5 years ago
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    because - -cos3x/3 = +cos3x/3 two negatives in math make a positive

  26. anonymous
    • 5 years ago
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    sin(x) was taken care of by substituting -du=sinxdx

  27. anonymous
    • 5 years ago
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    ohh is that where the (-1) came from? from -du = sinx?

  28. anonymous
    • 5 years ago
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    yeah there you go

  29. anonymous
    • 5 years ago
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    Then where did the cos go? from cos^2xsinx I get that u=cos and du = -sinx but what about the other cos?

  30. anonymous
    • 5 years ago
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    bc u=cos(x) \[\cos^2(x)= (\cos(x))^2\]

  31. anonymous
    • 5 years ago
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    Ohhh thank you sooo much to everyone who helped!! I get it now ! :D

  32. anonymous
    • 5 years ago
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    your welcome...... hope this helped

  33. anonymous
    • 5 years ago
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    I made a typing error again.... here is the whole thing without mistakes \[∫\sin^3(x)dx=∫\sin^2(x)\sin(x)dx\] \[∫(1−\cos^2(x))\sin(x)dx\] \[∫\sin(x)dx-∫\cos^2(x)\sin(x)dx\] \[u=\cos(x),du=−\sin(x)dx,−du=\sin(x)dx\] \[−\cos(x)+∫u^2du=−\cos(x)+1/3(\cos^3(x))+c\] \[\cos(x)[1/3(\cos^2(x))−1]+c\] sorry for the confusion

  34. anonymous
    • 5 years ago
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    let me know if you are lost

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