Results 1 to 7 of 7
Like Tree3Thanks
  • 1 Post By x3bnm
  • 1 Post By ILikeSerena
  • 1 Post By x3bnm

Math Help - Trigonometric Integrals Help

  1. #1
    Newbie
    Joined
    Feb 2013
    From
    Philippines
    Posts
    2

    Trigonometric Integrals Help

    Can you help me solve the integral of sin(x)/[cos^3 (x)] dx in three different ways?

    So far I have:a.) tanx sec^2x dx
    let u = tanx
    du = sec^2x dx
    = integral of u du
    = (u^2)/2
    = 1/2 tan^2 (x) + C

    b.) sinx/cos^3x dx
    (sinx)(cosx)^-3 dx
    let u = cos x
    du = -sinx
    = - integral of u^-3
    = 1/2 (cosx)^-2
    = 1/2 sec^2x + C

    I can't think of another way of solving it. Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: Trigonometric Integrals Help

    Quote Originally Posted by vanillachyrae View Post
    Can you help me solve the integral of sin(x)/[cos^3 (x)] dx in three different ways?

    So far I have:a.) tanx sec^2x dx
    let u = tanx
    du = sec^2x dx
    = integral of u du
    = (u^2)/2
    = 1/2 tan^2 (x) + C

    b.) sinx/cos^3x dx
    (sinx)(cosx)^-3 dx
    let u = cos x
    du = -sinx
    = - integral of u^-3
    = 1/2 (cosx)^-2
    = 1/2 sec^2x + C

    I can't think of another way of solving it. Thanks in advance!
    Hi vanillachyrae!

    How about u = sec x?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    300
    Thanks
    16

    Re: Trigonometric Integrals Help

    Also let u = \tan(x) and dv = \sec^2(x) then using integration by parts you can find the answer.

    Because we know that \int \frac{d}{dx}\left(u(x)v(x)\right)\,dx = \int u'(x)v(x)\,dx + \int u(x)v'(x)\,dx
    Last edited by x3bnm; March 17th 2013 at 05:40 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: Trigonometric Integrals Help

    Quote Originally Posted by x3bnm View Post
    Also let u = \tan^2(x) and dv = \sec^2(x) then using integration by parts you can find the answer.

    Because we know that \int \frac{d}{dx}\left(u(x)v(x)\right)\,dx = \int u'(x)v(x)\,dx + \int u(x)v'(x)\,dx
    Nice try... but...

    udv = \tan^2 x \sec^2 x dx = \frac{\sin^2x}{\cos^4x} \ne \frac{\sin x}{\cos^3x}
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    300
    Thanks
    16

    Re: Trigonometric Integrals Help

    Sorry ILikeSerena. In my last post I accidentally typed u = \tan^2(x) instead of u = \tan(x). I edited the last post just before your posting. Sorry for any confusion. Here's the complete proof.

    Let u = \tan(x) and dv = \sec^2(x)\,\,dx
    Now du = \sec^2(x)\,\,dx and v = \tan(x)

    So
    \begin{align*}\int \tan(x) \sec^2(x)\,\, dx =& \tan^2(x) - \int \tan(x) \sec^2(x)\,\,dx \\ =& \tan^2(x) - \frac{\tan^2(x)}{2} + C\\ =& \frac{\tan^2(x)}{2} + C \end{align*}
    Thanks from ILikeSerena
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member ILikeSerena's Avatar
    Joined
    Dec 2011
    Posts
    733
    Thanks
    121

    Re: Trigonometric Integrals Help

    Quote Originally Posted by x3bnm View Post
    Sorry ILikeSerena. In my last post I accidentally typed u = \tan^2(x) instead of u = \tan(x). I edited the last post just before your posting. Sorry for any confusion. Here's the complete proof.

    Let u = \tan(x) and dv = \sec^2(x)\,\,dx
    Now du = \sec^2(x)\,\,dx and v = \tan(x)

    So
    \begin{align*}\int \tan(x) \sec^2(x)\,\, dx =& \tan^2(x) - \int \tan(x) \sec^2(x)\,\,dx \\ =& \tan^2(x) - \frac{\tan^2(x)}{2} + C\\ =& \frac{\tan^2(x)}{2} + C \end{align*}
    Ah yes, that does work!
    But I'd do the proof like this.

    You have

    \int \tan(x) \sec^2(x)\,\, dx =& \tan^2(x) - \int \tan(x) \sec^2(x)\,\,dx

    Let I =\int \tan(x) \sec^2(x)\,\, dx
    Then

    \begin{align*}I =& \tan^2(x) - I \\ 2I =& \tan^2(x)\\ I =& \frac{\tan^2(x)}{2} + C \end{align*}
    Thanks from x3bnm
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member x3bnm's Avatar
    Joined
    Nov 2009
    Posts
    300
    Thanks
    16

    Re: Trigonometric Integrals Help

    Quote Originally Posted by ILikeSerena View Post
    Ah yes, that does work!
    But I'd do the proof like this.

    You have

    \int \tan(x) \sec^2(x)\,\, dx =& \tan^2(x) - \int \tan(x) \sec^2(x)\,\,dx

    Let I =\int \tan(x) \sec^2(x)\,\, dx
    Then

    \begin{align*}I =& \tan^2(x) - I \\ 2I =& \tan^2(x)\\ I =& \frac{\tan^2(x)}{2} + C \end{align*}
    Thank you ILikeSerena. Your method is neater than mine.
    Thanks from ILikeSerena
    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: January 20th 2013, 12:05 PM
  2. Trigonometric Integrals
    Posted in the Calculus Forum
    Replies: 3
    Last Post: November 14th 2009, 06:52 PM
  3. trigonometric integrals
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 28th 2009, 11:22 AM
  4. Trigonometric Integrals
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 23rd 2008, 03:08 PM
  5. Trigonometric Integrals
    Posted in the Calculus Forum
    Replies: 5
    Last Post: February 16th 2008, 05:18 PM

Search Tags


/mathhelpforum @mathhelpforum