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sat_chen

  • one year ago

Can someone explain me this trigonometric substitution

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  1. sat_chen
    • one year ago
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    http://www.wolframalpha.com/input/?i=integral+of+%281%2F%28x%5E3*sqrt%28x%5E2-25%29%29%29dx

  2. sat_chen
    • one year ago
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    so what i am confused is on how someone went from

  3. sat_chen
    • one year ago
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    |dw:1359567855994:dw|

  4. sat_chen
    • one year ago
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    to the solution

  5. sat_chen
    • one year ago
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    i think you have to log in to wolfram before you can see the steps

  6. sat_chen
    • one year ago
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    we get sin taking the integral of cos?

  7. shubhamsrg
    • one year ago
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    Yes I can't see the steps.

  8. shubhamsrg
    • one year ago
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    Was the substitution x = 5 siny ?

  9. sat_chen
    • one year ago
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    yea you have to log in for that i think and it would take me a year to type the solutions here XD

  10. sat_chen
    • one year ago
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    so someone with wolfram account tell me how they went from that step to the solution

  11. sat_chen
    • one year ago
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  12. shubhamsrg
    • one year ago
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    Leme try anyways, let x= 5secy => dx= 5 secy tany dy so we have inside integral 5 secy tany dy/( 125 sec^3 y ( -25 + 25sec^2 y)^1/2) => 5sec y tany dy/( 625 sec^3 y (tan^2 y)^1/2 ) => 1/125 ( dy/sec^2y)) or simply cos^2y dy Am I right so far ?

  13. sat_chen
    • one year ago
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    i got a better way for you to see the solutions this is part 2 of the solution

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  14. sat_chen
    • one year ago
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  15. sat_chen
    • one year ago
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    theres all the steps to the solution can you take me from that 0/250 + sin20/500 + c to the solution

  16. sat_chen
    • one year ago
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    and yes you are right with your steps

  17. sat_chen
    • one year ago
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    so from the integral of (1/125)(cos^20d0) you do the half angle formula to get integral of (1/125)(1+cos20)/2dp then you split and get 0/250 + sin2(theta)/500 + c so my question if how you get from here to the answer

  18. sat_chen
    • one year ago
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    |dw:1359570048133:dw|

  19. sat_chen
    • one year ago
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    i get the second part where it is the inverse of sec-1 (x/5) but i dont get where they got the 5sqrt(x^2) stuff from

  20. sat_chen
    • one year ago
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    |dw:1359570931896:dw| the stuff i dont get is in that box in case you are still confused on what i am confused about

  21. phi
    • one year ago
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    you are asking about \[ \frac{1}{2} \sin(2 u)\] with \[ u= \sec^{-1}\left(\frac{x}{5}\right) \] In terms of cosine, rather than secant, we can say \[ u= \cos^{-1}\left(\frac{5}{x}\right) \] If we take the hypotenuse to be x, in a triangle with side 5 |dw:1359572361980:dw| the other side is \( \sqrt{x^2-25} \) Using these sides, we have \[ \cos u = \frac{5}{x} \] and \[ \sin u = \frac{\sqrt{x^2-25}}{x} \] sin(2u) = 2 sin(u) cos(u) sub in the above definitions wolfram then multiplied by sqrt(x^2)/x (why?) to get your boxed expression

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