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Homework Help: 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]


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


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

    What would u do next??

  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


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    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


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    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]

    [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:


    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


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    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].

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