1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Homework Helper

  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


    User Avatar
    Homework Helper

    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:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook