Results 1 to 3 of 3

Math Help - Calc 2- Integration by Trigonmetric Substitution

  1. #1
    Newbie
    Joined
    Aug 2008
    Posts
    5

    Calc 2- Integration by Trigonmetric Substitution

    I was hoping someone could explain to me how to do the problem

    The integral from -pi/2 to 0 of cos(t)/(sqrt(1+sin^2(t)))dt


    if its possible will you please show every step because i am really bad at this!!!!



    Thank you
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    771
    Hello, fruitkate21!

    If you're "really bad at this", this problem is probably fatal . . .


    \int^0_{\text{-}\frac{\pi}{2}}\frac{\cos t\,dt}{\sqrt{1+\sin^2\!t}} . . This takes two subsitutions.

    Let: u \:=\:\sin t \quad\Rightarrow\quad du \:=\:\cos t\,dt

    Substitute: . \int\frac{du}{\sqrt{1+u^2}}


    Let: u = \tan\theta \quad\Rightarrow\quad du = \sec^2\theta\,d\theta \quad\Rightarrow\quad \sqrt{1+u^2} \:=\:\sec\theta

    Substitute: . \int\frac{\sec^2\!\theta\,d\theta}{\sec\theta} \;=\;\int\sec\theta\,d\theta \;=\;\ln|\sec\theta + \tan\theta|


    Back-substitute: . \tan\theta \:=\:\frac{u}{1} \:=\:\frac{opp}{adj} \quad\Rightarrow\quad hyp \:=\:\sqrt{1+u^2} \quad\Rightarrow\quad\sec\theta \:=\:\sqrt{1+u^2}

    . . and we have: . \ln|u + \sqrt{1+u^2}|


    Back-substitute: . u = \sin t

    . . and we have: . \ln\left|\,\sin t + \sqrt{1 + \sin^2\!t}\,\right|\:\bigg]^0_{\text{-}\frac{\pi}{2}}


    I'll let you evaluate it . . .

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by fruitkate21 View Post
    I was hoping someone could explain to me how to do the problem

    The integral from -pi/2 to 0 of cos(t)/(sqrt(1+sin^2(t)))dt


    if its possible will you please show every step because i am really bad at this!!!!



    Thank you
    First off, if you make the substitution z=\sin t, I leave it for you to show that the integral becomes \int_{-1}^0\frac{\,dz}{\sqrt{1+z^2}}

    Now apply the trig substitution z=\tan\theta\implies\,dz=\sec^2\theta\,d\theta Thus \int_{-1}^0\frac{\,dz}{\sqrt{1+z^2}}\xrightarrow{z=\tan\t  heta}\int_{-\frac{\pi}{4}}^0\frac{\sec^2\theta\,d\theta}{\sqrt  {1+\tan^2\theta}}=\int_{-\frac{\pi}{4}}^0\sec\theta\,d\theta.

    Now if you know how to evaluate \int\sec\theta\,d\theta, then see my next paragraph. I'll assume for now you don't know how to. Note that \int\sec\theta\,d\theta=\int\sec\theta\frac{\sec\t  heta+\tan\theta}{\sec\theta+\tan\theta}\,d\theta=\  int\frac{\sec^2\theta+\sec\theta\tan\theta}{\sec\t  heta+\tan\theta}\,d\theta. Now let \eta=\sec\theta+\tan\theta\implies\,d\eta=\left(\s  ec\theta\tan\theta+\sec^2\theta\right)\,d\theta. Thus, the integral becomes \int\frac{\,d\eta}{\eta}=\ln\!\left|\eta\right|+C=  \ln\!\left|\sec\theta+\tan\theta\right|+C.

    Since we were dealing with \int_{-\frac{\pi}{4}}^0\sec\theta\,d\theta, we see that this is equivalent to \left.\bigg[\ln\!\left|\sec\theta+\tan\theta\right|\bigg]\right|_{-\frac{\pi}{4}}^0. Thus, we have \ln\!\left|\sec\!\left(0\right)+\tan\!\left(0\righ  t)\right|-\ln\!\left|\sec\!\left(-\frac{\pi}{4}\right)+\tan\!\left(-\frac{\pi}{4}\right)\right|=\color{red}\boxed{-\ln\!\left(\sqrt{2}-1\right)}

    Hopefully, you were able to follow all of this. Does it make sense?

    EDIT: Soroban beat me to it...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Trig substitution calc 2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 1st 2010, 08:41 PM
  2. Calc integration using substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: January 26th 2009, 06:31 PM
  3. Calc: Substitution Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: December 3rd 2008, 09:44 AM
  4. Replies: 4
    Last Post: November 24th 2008, 06:14 AM
  5. integration using trigonmetric functions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 26th 2008, 04:12 PM

Search Tags


/mathhelpforum @mathhelpforum