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Definite integration

  1. May 24, 2013 #1
    1. The problem statement, all variables and given/known data
    Integration of sinx^4

    2. Relevant equations
    integration by parts

    3. The attempt at a solution
    I tried to break the function into 2 parts like (sin^3x)(sinx) and solve it using integration by parts but kept getting a function (sin^2x)(cos^2x) over and over again.
  2. jcsd
  3. May 24, 2013 #2
    Does it specifically ask for you to use integration by parts? Maybe using trig identities for ## \sin^2 (x) ## will be easier.

    If you get a recurring integral, try something like this:

    [tex]\displaystyle \int \sin^2 (x) \cos ^2 (x) = [\text{Integral stuff...}] - \int \sin^2 (x) \cos ^2 (x)[/tex]

    [tex]\displaystyle \int \sin^2 (x) \cos ^2 (x) + \int \sin^2 (x) \cos ^2 (x) = [\text{Integral stuff...}][/tex]

    [tex]\displaystyle 2 \int \sin^2 (x) \cos ^2 (x) = [\text{Integral stuff...}][/tex]

    [tex]\displaystyle \int \sin^2 (x) \cos ^2 (x) = \dfrac{[\text{Integral stuff...}]}{2}[/tex]
  4. May 24, 2013 #3
    It doesn't specifically say anything. I actually tried to use the trig identity sin^2x but got a very messy function to evaluate something like (1/2- 1/2cos2x)^2. I am posting what I actually got by applying integration by parts.

    ∫sin^3xsinx= sin^3xcosx- 3∫sin^2xcos^2x ............1
    =sin^3xcosx - 3[2sin^3xcosx -4∫cos^2xsin^2x]

    Let y= ∫sin^3xsinx
    from 1 we get, sin^3xcosx -3∫sin^2xcos^2x= sin^3xcosx-6sin^3xcosx+12∫cos^2xsin^2x
    → 6sin^3xcosx= 15∫sin^2xcos^2x
    → 5∫sin^2xcos^2x= 2sin^3xcosx
    → ∫sin^2xcos^2x=2sin^3xcosx/5+C
    Please check if there are any reckless mistakes and any step is wrong.
    Last edited: May 24, 2013
  5. May 24, 2013 #4


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    Science Advisor

    First, please note that "[itex]sin x^4[/itex], which is what you wrote, is NOT the same as "[itex]sin^4(x)[/itex] what is what I am sure was intended.

    The simplest way to integrate it is to use the fact that [itex]sin^4(x)= (1/2- (1/2)cos(2x))^2[/itex] as you say you started to use but apparently stopped at that point. Continuing, that is equal to [itex](1/4)- cos(2x)+ (1/4)cos^2(2x)= 1/4- cos(2x)+ (1/4)(1/2+ (1/2)cos(4x)= 3/8- cos(2x)+ 1/8 cos(4x)[/itex].

    Now as for your integration by parts, I have no idea what such things as
  6. May 24, 2013 #5

    Ray Vickson

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    Science Advisor
    Homework Helper

    If you know how to do ##F(t) = \int \sin^2 (t)\, dt##, then you can write
    [tex] \sin^4 (x) = \sin^2 (x)[1-\cos^2 (x)] = \sin^2 (x) - (\sin(x) \cos(x))^2 =
    \sin^2 (x) - \frac{1}{4} \sin^2 (2x)[/tex] and get the solution in terms of ##F(x)## and ##F(2x)##.
    Last edited: May 24, 2013
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