Results 1 to 6 of 6

Thread: Trigonometric Integrals

  1. #1
    Newbie
    Joined
    Feb 2008
    Posts
    21

    Trigonometric Integrals

    Can anyone help me or give me a good idea of how to approach the definite integral ([sinx]^2)*([cosx]^2) with the limits of integration from 0 to pi/2? I'm pretty sure I should somehow be able to simplify it using half-angle identities but I'm stuck. Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    $\displaystyle
    \sin ^2 x\cos ^2 x = \frac{1}
    {4}\sin ^2 2x = \frac{1}
    {4}\left( {\frac{{1 - \cos 4x}}
    {2}} \right) = \frac{{1 - \cos 4x}}
    {8}$

    $\displaystyle

    ^* \sin 2x = 2\sin x\cos x


    $
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    Use the identities $\displaystyle sin^{2}(x)=\frac{1-cos(2x)}{2}$ and $\displaystyle cos^{2}(x)=\frac{(1+cos(2x)}{2}$

    Then, $\displaystyle \frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-cos(2x))(1+cos(2x))dx$

    $\displaystyle =\frac{1}{4}\int_{0}^{\frac{\pi}{2}}(1-cos^{2}(2x))dx$$\displaystyle =\frac{1}{4}\int_{0}^{\frac{\pi}{2}}sin^{2}(2x)dx$$\displaystyle =\frac{1}{8}\int_{0}^{\frac{\pi}{2}}(1-cos(4x))dx$

    Now continue.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2008
    Posts
    21
    Is that an identity? That [1-cos(2x)^2] = sin(2x)^2 ? I'm not questioning you, I'm just making sure I understand what you did. Thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,002
    Thanks
    1
    Yes, that is an identity. Try deriving it.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    11,152
    Thanks
    731
    Awards
    1
    Quote Originally Posted by thejabronisayz View Post
    Is that an identity? That [1-cos(2x)^2] = sin(2x)^2 ? I'm not questioning you, I'm just making sure I understand what you did. Thanks
    You know that $\displaystyle 1 - cos^2(y) = sin^2(y)$ right? Well, let y = 2x ...

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trigonometric Integrals
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Apr 30th 2010, 03:46 AM
  2. Trigonometric Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Nov 14th 2009, 06:52 PM
  3. trigonometric integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 10th 2009, 07:45 AM
  4. Trigonometric Integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 22nd 2009, 03:14 AM
  5. Trigonometric Integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Oct 23rd 2008, 03:08 PM

Search Tags


/mathhelpforum @mathhelpforum