Results 1 to 8 of 8

Math Help - a couple of trigonometric integrals

  1. #1
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3

    a couple of trigonometric integrals

     \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx

     \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx

    My initial thought was to introduce a parameter and differentiate, but that approach didn't get me anywhere.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Random Variable View Post
     \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx

     \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx
    Here's how you do the first one:

    Spoiler:
    Let


    \displaystyle I=\int_0^\pi x\sin\left(\cos^2(x)\right)\cos^2(\sin(x))\text{ }dx


    Then, letting z=\pi-x we find that


    \displaystyle \begin{aligned}I &=\int_0^\pi (\pi-z)\sin\left(\cos^2(z)\right)\cos\left(\sin^2(z)\ri  ght)\text{ }dz\\ &=\pi\int_0^\pi\sin\left(\cos^2(z)\right)\cos\left  (\sin^2(z)\right)\text{ }dz-I\end{aligned}


    Thus,


    \displaystyle \frac{2}{\pi}I=\int_0^\pi \sin\left(\cos^2(z)\right)\sin\left(\cos^2(z)\righ  t)\text{ }dz


    Recall though that


    \displaystyle \sin(\phi)\cos(\theta)=\frac{\sin(\phi+\theta)+\si  n(\phi-\theta)}{2}


    And thus


    \displaystyle \begin{aligned}\frac{2}{\pi} I &= \int_0^\pi \sin\left(\cos^2(z)\right)\cos^2\left(\sin(z)\righ  t)\text{ }dz\\ &= \frac{1}{2}\int_0^\pi\left(\sin\left(\cos^2(z)+\si  n^2(z)\right)+\sin\left(\cos^2(z)-\sin^2(z)\right)\right)\text{ }dz\\ &=\frac{1}{2}\pi\sin(1)+\frac{1}{2}\int_0^\pi \sin(\cos(2z))\text{ }dz\end{aligned}


    Note though that


    \displaystyle \int_0^\pi\sin(\cos(2z))\text{ }dz=\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz+\int_{\frac{3\pi}{4}}^\pi\sin(\cos(2z))\text{ }dz+\int_{\frac{3\pi}{4}}^{\pi}\sin(\cos(2z))\text  { }dz


    Note though that by letting \displaystyle u=z-\frac{3\pi}{4} one gets that


    \displaystyle \int_{\frac{3\pi}{4}}^{\pi}\sin(\cos(2z))=\int_0^{  \frac{\pi}{4}}\sin(\cos(2u))\text{ }du


    And thus


    \displaystyle \int_0^\pi\sin(\cos(2z))\text{ }dz=2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz+\int_{\frac{\pi}{4}}^{\frac{3\pi}}{4}\sin(\cos  (2z))\text{ }dz


    But, letting w=z-\frac{\pi}{2} in this second integral one arrives at


    \displaystyle \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}\sin(\cos(2z)  )=-\int_{\frac{-\pi}{4}}^{\frac{\pi}{4}}\sin(\cos(2w))\text{ }dw


    and since the integrand is even this is equal to


    \displaystyle -2\int_0^{\frac{\pi}{4}}\sin(\cos(2w))\text{ }dw


    Therefore,


    \displaystyle \begin{aligned}\int_0^\pi\sin(\cos(2z))\text{ }dz &= 2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))+\int_{\frac{  \pi}{4}}^{\frac{3\pi}{4}}\sin(\cos(2z))\text{ }dz\\ &= 2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz-2\int_0^{\frac{\pi}{4}}\sin(\cos(2z))\text{ }dz\\ &=0\end{aligned}


    Thus, finally we arrive at


    \displaystyle \begin{aligned}\frac{2}{\pi}I &=\frac{\pi}{2}\sin(1)+\frac{1}{2}\int_0^\pi\sin(\  cos(2z))\text{ }dz\\ &=\frac{\pi}{2}\sin(1)\end{aligned}


    and thus \displaystyle I=\frac{\pi^2 \sin(1)}{4}.

    Last edited by Drexel28; December 26th 2010 at 08:54 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member Random Variable's Avatar
    Joined
    May 2009
    Posts
    959
    Thanks
    3
    The ODE you had was solvable.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Random Variable View Post
    The ODE you had was solvable.
    See my new method. It's more elementary.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Random Variable View Post
     \int_{0}^{\pi} x \sin (\cos^{2}x)\cos(\sin^{2}x) \ dx

     \int_{0}^{\pi} x \Big(\sin^{2}(\sin x) + \cos^{2} (\cos x) \Big)\ dx

    My initial thought was to introduce a parameter and differentiate, but that approach didn't get me anywhere.
    The second one is interesting. I have a solution, but it's a tad long. Do you have any ideas as of yet?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    to prove that \displaystyle\int_0^\pi\sin(\cos 2x)\,dx=0, split it up into two integrals with 0\le x\le\dfrac\pi2 and \dfrac\pi2\le x\le\pi then apply substitution t=x-\dfrac\pi2 on the second piece and you'll see is equal to minus the first piece, hence, the result.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    55
    Here's my solution to the second

    Spoiler:

    First use the double angle formula so

    \displaystyle \int_0^{\pi} x\left( \dfrac{1-\cos \left(2 \sin x\right)}{2} + \dfrac{1 + \cos \left(2 \cos x\right)}{2} \right)\, dx

    \displaystyle  = \int_0^{\pi} x\,dx + \frac{1}{2}\int_0^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\, dx

    We will show the second integral is zero by considering it in two parts. (**)

    First, that \displaystyle   \int_0^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\,dx = \int_{\pi/2}^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx

    Let x = \pi -u in \displaystyle   \int_{\pi/2}^{\pi} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx

    Next that \displaystyle  \int_0^{\pi/4} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\,dx = -\int_{\pi/4}^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)dx

    Let x = \pi/2-u in \displaystyle   \int_{\pi/4}^{\pi/2} \cos \left(2 \cos x\right) - \cos \left(2 \sin x\right)\, dx

    Thus, we are left with \displaystyle  \int_0^{\pi} x\,dx  = \dfrac{\pi^2}{2}

    Last edited by Jester; December 27th 2010 at 08:00 AM. Reason: Added new info
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    here's mine to the second:

    put t=\pi-x on the integral to see that is equal to \displaystyle\frac\pi2\int_0^\pi\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx.

    i claim the latter integral equals \pi, hence the original integral equals \dfrac{\pi^2}2, so in order to see that write the integral as

    \displaystyle\int_{0}^{\frac{\pi }{2}}{\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx}+\int_{\frac{\pi }{2}}^{\pi }{\left( {{\sin }^{2}}(\sin x)+{{\cos }^{2}}(\cos x) \right)\,dx},

    and put t=x-\dfrac\pi2 on the second one, then add them up and you'll see that the claimed integral achieve the aforesaid value, as required.

    another solution for the first one:

    we consider \displaystyle\int_{0}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}, (1) then write it as \displaystyle\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}+\int_{\frac{\pi }{2}}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx} and put t=x-\dfrac\pi2 on the second one then we get \displaystyle\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}+\int_{0}^{\frac{\pi }{2}}{\sin \left( {{\cos }^{2}}x \right)\cos \left( {{\sin }^{2}}x \right)\,dx}, so when adding those we note that the integrand is actually \sin(\sin^2x+\cos^2x) so (1) equals \dfrac\pi2\sin1.

    on the original integral was easy to show that it's equal to \displaystyle\frac\pi2\int_{0}^{\pi }{\sin \left( {{\sin }^{2}}x \right)\cos \left( {{\cos }^{2}}x \right)\,dx}, thus by (1) we conclude.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 14
    Last Post: April 6th 2011, 11:22 AM
  2. a couple more definite integrals
    Posted in the Math Challenge Problems Forum
    Replies: 2
    Last Post: March 31st 2011, 09:44 AM
  3. Need help with couple of integrals!
    Posted in the Calculus Forum
    Replies: 9
    Last Post: June 1st 2010, 11:32 AM
  4. a couple integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: April 18th 2008, 01:20 PM
  5. A couple of indefinite integrals
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 18th 2006, 01:29 AM

Search Tags


/mathhelpforum @mathhelpforum