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!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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
    mean.
     
  6. May 24, 2013 #5

    Ray Vickson

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

Have something to add?
Draft saved Draft deleted



Similar Discussions: Definite integration
  1. Definite Integral (Replies: 7)

  2. Definite Integral (Replies: 10)

  3. Definite integral (Replies: 4)

  4. Definite integral (Replies: 6)

Loading...