Results 1 to 3 of 3

Math Help - Integral Help

  1. #1
    Junior Member
    Joined
    Oct 2012
    From
    London
    Posts
    54

    Integral Help

    Integral Help-codecogseqn.gif
    I am stuck on this problem. I tried substituting u= cos^3(x^2), but it didnt work for me.
    I am not asking for someone to do this for me I just need some help getting started.
    Thanks in advance.
    Last edited by nubshat; November 22nd 2012 at 07:37 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    12,125
    Thanks
    1009

    Re: Integral Help

    Quote Originally Posted by nubshat View Post
    Click image for larger version. 

Name:	CodeCogsEqn.gif 
Views:	3 
Size:	1.3 KB 
ID:	25840
    I am stuck on this problem. I tried substituting u= cos^3(x^2), but it didnt work for me.
    I am not asking for osmeone to do this ofr me I just need soem help getting started.
    Thanks in advance.
    u = x^2

    du = 2x \, dx

    \frac{1}{2} \int_0^{\sqrt{\frac{\pi}{6}}} \cos(x^2)[\cos^2(x^2) - 1] \cdot 2x \, dx

    -\frac{1}{2} \int_0^\frac{\pi}{6} \cos{u}(\sin^2{u}) \, du

    t = \sin{u}

    dt = \cos{u} \, du

    -\frac{1}{2} \int_0^\frac{1}{2} t^2 \, dt
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,911
    Thanks
    773

    Re: Integral Help

    Hello, nubshat!

    \displaystyle \int^{\sqrt{\frac{\pi}{6}}}_0\left[\cos^3(x^2) - \cos(x^2)\right]x\,dx

    Note that: . \cos^3(x^2) - \cos(x^2) \;=\;\cos(x^2)\left[\cos^2(x^2) - 1\right] \;=\;\cos(x^2)\left[-\sin^2(x^2)\right]

    The integral becomes: . \displaystyle -\int\sin^2(x^2)\cdot \cos(x^2)\,x\,dx

    Let u = \sin(x^2) \quad\Rightarrow\quad du = \cos(x^2)\cdot2x\cdot dx \quad\Rightarrow\quad \cos(x^2)\,x\,dx = \tfrac{1}{2}du
    . . When x = 0\!:\;u = \sin(0^2) = 0.
    . . When x = \sqrt{\tfrac{\pi}{6}}\!:\;u = \sin\tfrac{\pi}{6} = \tfrac{1}{2}

    Substitute: . \displaystyle -\int^{\frac{1}{2}}_0 u^2\cdot\tfrac{1}{2}du \;=\;-\tfrac{1}{2}\int^{\frac{1}{2}}_0 u^2du \;=\;-\tfrac{1}{6}u^3\,\bigg]^{\frac{1}{2}}_0

    . . . . . . . . =\;\left[-\tfrac{1}{6}\left(\tfrac{1}{2}\right)^3\right] - \left[-\tfrac{1}{6}(0^3)\right] \;=\;-\tfrac{1}{48}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 08:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 03:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 02:52 PM
  4. [SOLVED] Line integral, Cauchy's integral formula
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: September 16th 2009, 12:50 PM
  5. Replies: 0
    Last Post: September 10th 2008, 08:53 PM

Search Tags


/mathhelpforum @mathhelpforum