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Trigonometric Substitution

  1. Apr 7, 2005 #1
    [tex]\int x^3\sqrt{4-9x^2}dx[/tex]

    I tried to use [tex]x=\frac{2}{3}\cos{(x)}[/tex] but it just left me with [tex]\int \sin^3{(x)}\cos^2{(x)}dx[/tex]

    Any suggestions?

    Thanks for your help.
     
  2. jcsd
  3. Apr 7, 2005 #2

    dextercioby

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    It's the oher way around.U missed a numerical factor (containing an essential minus)...U need to substitute

    [tex] x=\frac{2}{3}\sin u [/tex]

    to get the powers of sine & cosine that u have there.U'd still miss the numerical factor and but at least you'd have gotten rid of the "-"...

    Let's assume this one (it's your transformed integral up to a numerical factor).

    [tex] \int \sin^{3}u \cos^{2} u \ du [/tex]...

    Use the fundamental identity of circular trigonometry to write it

    [tex]\int \sin^{3}u \ du -\int \sin^{5}u \ du [/tex]

    The first can be written as

    [tex]-\int (1-\cos^{2}u) \ d(\cos u) [/tex]

    The second (the integral,without the "-" preceding it).

    [tex] -\int \left(1-\cos^{2}u\right)^{2} \ d(\cos u) [/tex]

    ,which are very easy to integrate...


    Daniel.
     
    Last edited: Apr 7, 2005
  4. Apr 7, 2005 #3
    [tex]\int{\sin^3{\theta}}d\theta - \int{\sin^5{\theta}}d\theta[/tex]

    I know how to integrate the first, but the second I am having a bit more trouble with.

    Is this true?: [tex]\int{\sin^5{\theta}}d\theta = -\int{(1-\cos{\theta})^2}d\cos{\theta}[/tex]

    Thanks again.
     
  5. Apr 7, 2005 #4

    dextercioby

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    I told u what it is equal to.

    [tex] \int \sin^{5}x \ dx=-\int \left(1-\cos^{2}x\right)^{2} d(\cos x) [/tex]

    Daniel.
     
  6. Apr 7, 2005 #5
    I'm sorry, what I meant was:

    What is the best way to integrate it? I haven't seen one with the square before, so I'm not sure...

    Thanks for your help.
     
  7. Apr 7, 2005 #6

    HallsofIvy

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    If you are going to use "trigonometric substitution" you should learn to integrate trig functions first! :smile:

    Any time you have a an odd power of sin or cos, factor out one of then to go with the "dx", then convert the even power using sin2x+ cos2x= 1.

    For example, seeing [tex]\int sin^5 x dx[/tex], I thinkof it as [tex]\int sin^4 x (sin x dx)= \int (sin^2 x)^2 (sin x dx)= \int (1- cos^2 x)(sin x dx)[/tex]. Now let u= cos(x) so that du= -sin x dx and that becomes [tex]\int (1-u^2)^2 du[/tex].

    If you have only even powers of sin and cos, you can't do that- you have to use the trig identities [tex] cos^2(x)= \frac{1}{2}[1+ cos(2x)] [/tex] and [tex]sin^2(x)= \frac{1}{2}[1- cos(2x)][/tex] to reduce the powers.
     
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