Results 1 to 6 of 6

Thread: Indefinite Trig Integrals: int (sin^4x)(cos^2x) dx and

  1. #1
    Junior Member
    Join Date
    Sep 2005
    Location
    toronto
    Posts
    93

    Indefinite Trig Integrals: int (sin^4x)(cos^2x) dx and

    I am doing a chapter on trigonometic integrals. I have just come across a couple questions that I don't know how to rearrange at the beginning to solve for the integral..

    1. integral of (sin^4x)(cos^2x) dx

    Because the powers are both even, the book suggests using half angle formulas.


    2. integral of (sin^3x)(sqrt cosx) dx

    If someone could just help me with the first step of rearranging in order to solve that would be great!
    thank you!!

  2. #2
    Senior Member skeeter's Avatar
    Join Date
    Dec 2005
    Location
    Fort Worth, TX
    Posts
    2,406
    [tex]\L \sin^4{x}\cos^2{x} =[/tex]

    [tex]\L (\sin^2{x})^2\cos^2{x} =[/tex]

    [tex]\L \left(\frac{1-\cos{(2x)}}{2}\right)^2 \left(\frac{1 + \cos{(2x)}}{2}\right) =[/tex]

    [tex]\L \left(\frac{1 - cos^2{(2x)}}{4}\right) \left(\frac{1 - \cos{(2x)}}{2}\right)[/tex]

    [tex]\L \left(\frac{\sin^2{(2x)}}{4}\right) \left(\frac{1 - \cos{(2x)}}{2}\right) =[/tex]

    [tex]\L \frac{1}{8} \left[\sin^2{(2x)} - \sin^2{(2x)}\cos{(2x)}\right] =[/tex]

    [tex]\L \frac{1}{8} \left[\frac{1 - \cos{(4x)}}{2} - \sin^2{(2x)}\cos{(2x)}\right] =[/tex]

    you now should be able to integrate the last expression.



    for the 2nd one ...

    [tex]\L \sin^3{x} \cdot \sqrt{\cos{x}} =[/tex]

    [tex]\L \sin{x}(1 - cos^2{x})\sqrt{\cos{x}} =[/tex]

    [tex]\L \left(\cos^{\frac{5}{2}}{x} - \cos^{\frac{1}{2}}{x}\right)(-\sin{x})[/tex]

    finish up.

  3. #3
    Junior Member
    Join Date
    Sep 2005
    Location
    toronto
    Posts
    93
    how did you get from this step to the next?

    from here [tex]\L \left(\frac{\sin^2{(2x)}}{4}\right) \left(\frac{1 - \cos{(2x)}}{2}\right) =[/tex]

    to here [tex]\L \frac{1}{8} \left[\sin^2{(2x)} - \sin^2{(2x)}\cos{(2x)}\right] =[/tex]


    .. and thank you.. i was able to figure out the 2nd one!
    thank you!!

  4. #4
    Senior Member skeeter's Avatar
    Join Date
    Dec 2005
    Location
    Fort Worth, TX
    Posts
    2,406
    Quote Originally Posted by maeveoneill
    how did you get from this step to the next?

    from here [tex]\L \left(\frac{\sin^2{(2x)}}{4}\right) \left(\frac{1 - \cos{(2x)}}{2}\right) =[/tex]

    to here [tex]\L \frac{1}{8} \left[\sin^2{(2x)} - \sin^2{(2x)}\cos{(2x)}\right] =[/tex]

    no big mystery ... I factored out the two constants in the denominator and multiplied (distributed) the numerators ...
    sin<sup>2</sup>(2x)[1 - cos(2x)] = sin<sup>2</sup>(2x) - sin<sup>2</sup>(2x)cos(2x)



    .. and thank you.. i was able to figure out the 2nd one!

  5. #5
    Junior Member
    Join Date
    Sep 2005
    Location
    toronto
    Posts
    93
    then you would use substitution, u = sin2x and du= 1/2cos2x dx
    would your du not cancel out the cos 2x in the equation and then elave you with u^2 - u^2?? = 0?
    thank you!!

  6. #6
    Senior Member skeeter's Avatar
    Join Date
    Dec 2005
    Location
    Fort Worth, TX
    Posts
    2,406
    [tex]\L \frac{1}{8}\int \left[\sin^2{(2x)} - \sin^2{(2x)}\cos{(2x)}\right] dx[/tex]

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

    [tex]\L \frac{1}{8} \int \frac{1 - \cos{(4x)}}{2} dx - \frac{1}{16}\int \sin^2{(2x)} \cdot 2\cos{(2x)}dx[/tex]

    [tex]\L \frac{1}{16} \int 1 - \cos{(4x)} dx - \frac{1}{16}\int u^2 du[/tex]

    [tex]\L \frac{1}{16}\left[x - \frac{\sin{(4x)}}{4} - \frac{\sin^3{(2x)}}{3}\right] + C[/tex]

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •