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anonymous

  • 5 years ago

hey can someone please help me find the integral of sqrt of y^2-49. i know i have to use trig substitution but i need help with actually solving it

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  1. myininaya
    • 5 years ago
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    \[\int\limits_{}^{}\sqrt{y^2-49}dy\] let \[\frac{y}{7}=\tan(\theta)\] \[\frac{dy}{7}=\sec^2\theta d \theta\] \[\int\limits_{}^{}\sqrt{(7\tan \theta)^2-49}7\sec^2\theta d \theta\]

  2. anonymous
    • 5 years ago
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    isnt it better to use 7 sin theta as a sunstitution

  3. myininaya
    • 5 years ago
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    \[7\int\limits_{}^{}\sqrt{49}\sqrt{\tan^2\theta-1}\sec^2(\theta)d \theta\]

  4. myininaya
    • 5 years ago
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    \[7*7\int\limits_{}^{}\sec \theta*\sec^2(\theta) d \theta\]

  5. myininaya
    • 5 years ago
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    \[49\int\limits_{}^{}\sec^3(\theta) d \theta\]

  6. myininaya
    • 5 years ago
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    oops i made the wrong substitution

  7. anonymous
    • 5 years ago
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    yah lol

  8. myininaya
    • 5 years ago
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    tan^2x+1=sec^2x not tan^2x-1=sec^2x

  9. anonymous
    • 5 years ago
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    could you do it using sin theta as the substitution if it not too much trouble

  10. anonymous
    • 5 years ago
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    this is the part im stuck if it helps

  11. myininaya
    • 5 years ago
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    you could also use sintheta=y/7

  12. anonymous
    • 5 years ago
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    i got 49 cos^2 theta

  13. myininaya
    • 5 years ago
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    i mean costhetha=y/7

  14. anonymous
    • 5 years ago
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    then i knew i i got the integral of cos^2 theta which came out to ve 1/2 theta -1/4 cos 2 theta

  15. myininaya
    • 5 years ago
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    \[\sin \theta=\frac{y}{7}\] \[\cos(\theta)d \theta=\frac{dy}{7}\] \[(\sin \theta)^2=(\frac{y}{7})^2\] \[49\sin^2(\theta)=y^2\] \[\int\limits_{}^{}\sqrt{49\sin^2(\theta)-49}*7\cos(\theta)d \theta\] \[7\int\limits_{}^{}\sqrt{49}*\sqrt{\sin^2(\theta)-1}*\cos(\theta)d \theta=49\int\limits_{}^{}\cos^2\theta d \theta\]

  16. myininaya
    • 5 years ago
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    remember cos^2(theta)=1/2*(1+cos(2theta))

  17. myininaya
    • 5 years ago
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    \[49*\frac{1}{2}\int\limits_{}^{}(1+\cos(2\theta)) d \theta=\frac{49}{2}*(\theta+\frac{1}{2}\sin (2\theta))+C\]

  18. myininaya
    • 5 years ago
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    but remember sin(2theta)=2sin(theta)cos(theta)

  19. myininaya
    • 5 years ago
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    \[\frac{49}{2} \theta+\frac{49}{2} \sin(\theta)\cos(\theta)+C\]

  20. myininaya
    • 5 years ago
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    we need this in terms of x though we let sintheta=y/7 so costhetha=sqrt(49-y^2)/7

  21. myininaya
    • 5 years ago
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    also since sintheta=y/7 then theta=arcsin(y/7)

  22. myininaya
    • 5 years ago
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    meant in terms of y lol

  23. anonymous
    • 5 years ago
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    hey i forgot to give you the limits bc i am having trouble plugging it its 0 to 7

  24. myininaya
    • 5 years ago
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    \[\frac{49}{2}\sin^{-1} (y)+\frac{49}{2}\frac{y}{7}\frac{\sqrt{49-y^2}}{7}+C\]

  25. myininaya
    • 5 years ago
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    \[\frac{49}{2}\sin^{-1} (y)+\frac{y}{2}\sqrt{49-y^2}+C \]

  26. myininaya
    • 5 years ago
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    you are having trouble plugging in?

  27. myininaya
    • 5 years ago
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    wherever you see a y plug in your upper limit then minus wherever you see a y plug in your lower limit

  28. myininaya
    • 5 years ago
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    \[(\frac{49}{2}\sin^{-1}(7)+\frac{7}{2}\sqrt{49-7^2})-(\frac{49}{2}\sin^{-1}(0)+\frac{0}{2}\sqrt{49-0^2})\]

  29. myininaya
    • 5 years ago
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    \[\frac{49}{2}\sin^{-1}(7)\]

  30. anonymous
    • 5 years ago
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    ty :)

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