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Homework Help: Trigonometric substitution integrals

  1. Sep 12, 2004 #1
    what would be the appropriate trigonometic substition for the following:

    1.) [tex]\int(7x^2-7)^{3/2}) dx [/tex]

    2.) [tex] \int \frac{x^2}{\sqrt{(7x^2 + 8)}} [/tex]

    ok i have a book with a list of trigonometric functions in front of me, but i dont get it. i dont see anything that i can sub in for.

    ok i kinda get the first one...


    [tex]sin^2(\theta) + cos^2(\theta) = 1/[/tex] so...
    divided by [tex]cos^2(\theta)[/tex] and get [tex]tan^2(\theta)[/tex]

    [tex]7(x^2 - tan^2(\theta))^{3/2}[/tex]

    [tex]7(x^3 - tan^3(\theta))[/tex]

    [tex]7x^3 - 7tan^3(\theta)[/tex]

    divide by 7.... (lol, dont think i can do this)

    [tex]x^3 = tan^3(\theta)[/tex]

    cube root...

    [tex]x= tan(\theta)[/tex] is that right?
    Last edited: Sep 12, 2004
  2. jcsd
  3. Sep 12, 2004 #2


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  4. Sep 12, 2004 #3
    lol, thanks. i just made a dumb mistake
  5. Sep 12, 2004 #4
    [tex] \int \frac{x^2}{\sqrt{(7x^2 + 8)}} [/tex]

    ok this one is a bit hard....i been thinking about this one for awhile.

    can i do [tex] \int \frac{x^2}{\sqrt{(7x^2 + 7 + 1)}} [/tex]

    [tex]sin^2(\theta) + cos^2(\theta) = 1[/tex] so...

    well i know it's [tex]sec^2(\theta)[/tex] from solving the previous problem

    [tex] \int \frac{x^2}{\sqrt{(7x^2 + 7 + sec^2(\theta))}} [/tex]

    [tex] \int \frac{x^2}{\sqrt{7(x^2 + 1 + sec^2(\theta))}} [/tex]

    agian, there's another 1! so i use [tex]sec^2(\theta)[/tex]
    (just working with the bottom part now)

    [tex] \sqrt{7(x^2 + sec^2(\theta) + sec^2(\theta)[/tex]

    if i squared 7, it would be a mess, can someone help?
    Last edited: Sep 12, 2004
  6. Sep 12, 2004 #5
    You might want to review trigonometric substitution in your textbook.
  7. Sep 12, 2004 #6


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    Check what are you doing again, it seems a bit... mixed up...

    Here's a hint

    [tex] \sqrt{7}x = \sqrt{8}tan\theta [/tex]

    so you should have

    [tex] \sqrt{7x^2 + 8} = \sqrt{8}sec\theta [/tex]
  8. Sep 12, 2004 #7
    i get it now! thanks.

    the trig. sub. should be:

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