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Integration Using Trigonometric Substitution Help Needed!

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  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data
    Integral of $$ x^3\sqrt{x^2+16}dx $$
    answer should give
    $$ 1/5(x^2+16)^{5/2} -16/3(x^2+16)^{1/2}+C $$
    2. Relevant equations
    x=atanθ

    3. The attempt at a solution

    IMG_20150222_141431.jpg
    Mod note: The integral is ##\int x^3 \sqrt{x^2 + 16} dx##
    The published answer is ##1/5(x^2+16)^{5/2} -16/3(x^2+16)^{1/2}+C##

    To the OP: Using $ $ (i.e., with a space between the $ symbols) causes the browser to not render what's between the $ pairs. For inline LaTeX, you can use # # (no space) at the beginning and end.
     

    Attached Files:

    Last edited: Feb 22, 2015
  2. jcsd
  3. Feb 22, 2015 #2
    Please explain what you are trying to do
     
  4. Feb 22, 2015 #3
    I'm trying to find the integral of the equation, but it doesn't give the same answer as the one I'm supposed to get, and I don't know what I'm doing wrong. I'm using trigonometric substitution which lets x=4sinΘ.
     
  5. Feb 22, 2015 #4
    In the second step after you did the substitution, the whole things becomes,
    4*((tan0^2)+1)^(1/2)=4sec0 and not 4sec0^2
     
  6. Feb 22, 2015 #5
    can you post the question? Like the integral you are trying to find the answer for?
     
  7. Feb 22, 2015 #6
    IMG_20150222_143359~2.jpg

    No problem! I tried using Latex, but it doesn't work...It's #6
     

    Attached Files:

  8. Feb 22, 2015 #7
    I might be able to help you....I love integration...but I don't understand anything from the picture you posted
     
  9. Feb 22, 2015 #8

    Mark44

    Staff: Mentor

    See my previous post. It's fairly clear which integral the OP is trying to find.
     
  10. Feb 22, 2015 #9

    Mark44

    Staff: Mentor

    See my edit to post #1.
     
  11. Feb 22, 2015 #10
    Thank You Mark! I'm still learning to use Latex... I changed it and it should appear ok, now!
     
  12. Feb 22, 2015 #11
  13. Feb 22, 2015 #12
    Ok, I'm currently trying, correcting the mistake I made with my sec^2. It looks good!
     
  14. Feb 22, 2015 #13
    I can officially say that it works now! Thank you Faris! Such a "minor" mistake that really changed everything in the problem!
     
  15. Feb 22, 2015 #14
    He meant Question 6 from the picture
     
  16. Feb 22, 2015 #15
    Your trig substitution is correct. When you square root, you should get 4sec(x) (I'll use x instead of theta for simplicity)
    That will leave you with,
    64tan3(x) * 4sec(x)

    or simply
    256tan3(x)*sec(x)

    Break the tan3(x) up, you get
    256tan2(x)*sec(x)*tan(x)

    I wont do the rest for you but I'll give you a clue. It involves a trig identity and a u substitution
     
  17. Feb 22, 2015 #16
    Thank you! It works now, my mistake was simply that I forgot to take the sqrt of sec^2
     
  18. Feb 22, 2015 #17
    hey, if trigonometric substitution is not necessary, it is much easier to put x^2+16=t.......=>x^2=t-16.......dx=dt/2x.......x^3=x*x2.....
    then integral is ##\int { x*(t-16)*\sqrt(t)*dt/2x} ## => ##\int {(t-16)\sqrt(t) dt/2} ## ............now expand normally...
     
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