Results 1 to 5 of 5

Math Help - Another Trig Integration by Parts Problem

  1. #1
    Senior Member
    Joined
    Oct 2012
    From
    USA
    Posts
    410
    Thanks
    1

    Another Trig Integration by Parts Problem

    \int9x\cos8x = [9x][\dfrac{1}{8}\sin x] - \int [\dfrac{1}{8}\sin x][9]

    u = 9x

    dv = \cos 8x dx

    v = \dfrac{1}{8}\sin x

    du = 9 dx

    \dfrac{9}{8}\sin^{2}x - \int\dfrac{9}{8} \sin x

    \dfrac{9}{8}\sin^{2}x + \dfrac{9}{8} \cos x - Online homework says this is wrong. ??
    Last edited by Jason76; June 29th 2014 at 10:28 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,557
    Thanks
    1424

    Re: Another Trig Integration by Parts Problem

    $\displaystyle \begin{align*} v = \frac{1}{8}\sin{(8x)} \end{align*}$, not $\displaystyle \begin{align*} \frac{1}{8}\sin{(x)} \end{align*}$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Oct 2012
    From
    USA
    Posts
    410
    Thanks
    1

    Re: Another Trig Integration by Parts Problem

    \int9x\cos8x = [9x][\dfrac{1}{8}\sin 8x] - \int [\dfrac{1}{8}\sin 8x][9] dx

    u = 9x

    dv = \cos 8x dx

    v = \dfrac{1}{8}\sin 8x

    du = 9 dx

    = [\dfrac{9}{8}\sin^{2} 8x] - \int [\dfrac{9}{8}\sin 8x]

    = [\dfrac{9}{8}\sin^{2} 8x] + [\dfrac{9}{64}\cos 8x] + C ???
    Last edited by Jason76; June 29th 2014 at 10:57 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,557
    Thanks
    1424

    Re: Another Trig Integration by Parts Problem

    NO! $\displaystyle \begin{align*} 9x\cdot \frac{1}{8}\sin{(8x)} \end{align*}$ is NOT $\displaystyle \begin{align*} \frac{9}{8}\sin^2{(8x)} \end{align*}$, it's $\displaystyle \begin{align*} \frac{9}{8}x\sin{(8x)} \end{align*}$.

    Also $\displaystyle \begin{align*} \int{\frac{9}{8}\sin{(8x)}\,\mathrm{d}x} = -\frac{1}{8}\cdot \frac{9}{8}\cos{(8x)} = -\frac{9}{64}\cos{(8x)} \end{align*}$, NOT $\displaystyle \begin{align*} \frac{72}{8}\cos{(8x)} \end{align*}$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Oct 2012
    From
    USA
    Posts
    410
    Thanks
    1

    Re: Another Trig Integration by Parts Problem

    The answer came out right on homework using ProveIt's suggestion.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig Integration by Parts Problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: July 2nd 2014, 09:11 PM
  2. Trig Sub to Integration by Parts (Supposedly)
    Posted in the Calculus Forum
    Replies: 6
    Last Post: October 8th 2010, 08:59 PM
  3. trig substitution and integration by parts
    Posted in the Trigonometry Forum
    Replies: 7
    Last Post: January 14th 2009, 08:01 PM
  4. Integration by parts and with trig
    Posted in the Calculus Forum
    Replies: 11
    Last Post: October 5th 2008, 07:27 PM
  5. Replies: 3
    Last Post: September 2nd 2008, 07:22 PM

Search Tags


/mathhelpforum @mathhelpforum