Results 1 to 3 of 3

Thread: 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
    12,028
    Thanks
    848
    Hello, fruitkate21!

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


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

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

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


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

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


    Back-substitute: .$\displaystyle \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: .$\displaystyle \ln|u + \sqrt{1+u^2}| $


    Back-substitute: .$\displaystyle u = \sin t$

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


    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
    5
    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 $\displaystyle z=\sin t$, I leave it for you to show that the integral becomes $\displaystyle \int_{-1}^0\frac{\,dz}{\sqrt{1+z^2}}$

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

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

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

    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: Oct 1st 2010, 07:41 PM
  2. Calc integration using substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Jan 26th 2009, 05:31 PM
  3. Calc: Substitution Rule
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Dec 3rd 2008, 08:44 AM
  4. Replies: 4
    Last Post: Nov 24th 2008, 05:14 AM
  5. integration using trigonmetric functions
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Oct 26th 2008, 03:12 PM

Search Tags


/mathhelpforum @mathhelpforum