1. Not finding help here? Sign up for a free 30min 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!

I hate Trigonometric Subsitution

  1. Jan 3, 2005 #1
    How would you do

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


    after you get to the point where you have made the triangle and you have

    [tex] x=sin\theta [/tex]

    and

    [tex] dx=cos\theta [/tex]
     
  2. jcsd
  3. Jan 3, 2005 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Yeah,Tom,with the substitution it becomes
    [tex] \int \cos^{4}x dx [/tex]

    What would u do next??

    Daniel.
     
  4. Jan 3, 2005 #3
    thats the problem... lol
    I was able to get to that point... I just don't know how to get to the answer... I checked it on my calcutor and found a really long answer.
     
  5. Jan 3, 2005 #4

    Pyrrhus

    User Avatar
    Homework Helper

    Tom, how about

    [tex] \int \cos^{4}x dx [/tex]

    [tex] \int (\cos^{2}x)^{2} dx [/tex]

    [tex] \int ( \frac{1 + \cos 2x}{2})^{2} dx [/tex]

    Edit: It was redudant, This is better, Yes dextercioby, i fixed it now, sorry for that :tongue2:
     
    Last edited: Jan 3, 2005
  6. Jan 3, 2005 #5

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Call your integral 'I'.I'm sure u still remember that:
    [tex]\cos^{4}x=\cos^{2}(1-\sin^{2}x) [/tex]
    Then 'I' is split into 2 integrals:
    [tex] I=\int (\cos^{2}x -\cos^{2}x\sin^{2}x) dx [/tex]
    Call the first integral by [itex]I_{1} [/itex]
    [tex] I_{1}=\int \cos^{2} x dx [/tex]

    HINT:[tex] \cos^{2}x=\frac{1}{2}(1+\cos 2x) [/tex]

    Call the second integral by [itex] I_{2} [/itex]
    [tex] I_{2}=\int \cos^{2}x\sin^{2}x dx [/tex]

    HINT:[tex]\sin 2x=2\sin x\cos x [/tex]
    [tex] \sin^{2} 2x=\frac{1}{2}(1-\cos 4x) [/tex]

    Then
    [tex] I=I_{1}-I_{2} [/tex]

    To return to the initial variable (remember u made the substitution x->sin x),simply make the 'x' in your result for 'I' go to [itex] \arcsin x [/itex] (i.e.reverse the substitution).

    Good luck!!I'm gonna go to bed now.By the time i get up,u'd better have it done... :wink:

    Daniel.

    EDIT:Cyclovenom,don't give him bad advice... :grumpy:
     
    Last edited: Jan 3, 2005
  7. Jan 4, 2005 #6
    ...

    sorry didn't finish it... or the rest of the homework

    I got Iintegral one down to

    [tex] 1/2 \int (x + \frac{sin2x}{2} ) [/tex]

    I started

    I2 but I wasn't sure where the [tex]sin^22x[/tex] hint came in since it was just sin

    Anyway I prob just didn't spend ennuogh time on it... I willl actually work on it some before that class...
     
  8. Jan 4, 2005 #7

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Okay,let's check it:
    [tex] I_{1}=\frac{1}{2}\int (1+\cos 2x) dx=\frac{1}{2}(x+\frac{1}{2}\sin 2x)=\frac{x}{2}+\frac{\sin 2x}{4} [/tex](1)

    Yous still has the integral sign.Maybe it was a typo...

    [tex] I_{2}=\int \sin^{2}x\cos^{2}x dx=\frac{1}{4}\int \sin^{2} 2x =\frac{1}{8}\int (1-\cos 4x) dx=\frac{1}{8}(x-\frac{1}{4}\sin 4x)=\frac{x}{8}-\frac{\sin 4x}{32}[/tex] (2)

    Now 'I' is the difference between the first intergral and the second,viz.
    [tex] I=(\frac{x}{2}-\frac{x}{8})+\frac{\sin 2x}{4}+\frac{\sin 4x}{32} =\frac{3x}{8}+\frac{\sin 2x}{4}+\frac{\sin 4x}{32} [/tex] (3)

    I'll let u invert the substitution [itex] x\rightarrow \arcsin x [/itex].

    Daniel.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: I hate Trigonometric Subsitution
  1. I hate pendulums (Replies: 11)

Loading...