Results 1 to 3 of 3

Thread: Integration using substitution

  1. #1
    Member
    Joined
    Mar 2007
    From
    England
    Posts
    104

    Integration using substitution

    I'm stuck on the following question:

    Use the subsitution t=sinhx to find:

    $\displaystyle
    \
    \int_0^a {\sqrt {(1 + t} ^2 )dt}
    \
    $

    I used
    $\displaystyle
    \
    \frac{{dt}}{{dx}} = \cosh x
    \
    $ and that
    $\displaystyle
    \
    1 + \sinh ^2 x= \cosh ^2 x
    \
    $

    So the integral becomes
    $\displaystyle
    \
    \int_0^{ar\sinh 1} {\cosh ^2 } xdx
    \
    $ and using the doule angle formulae I get
    $\displaystyle
    \
    \frac{1}{2}\int_0^{ar\sinh 1} {\cosh } 2x + 1dx
    \$
    Giving:
    $\displaystyle
    \
    \frac{1}{4}\sinh 2x + \frac{1}{2}x
    \$

    but then I can go no further because the answer had ln in and I don't know where that came from. Can someone help me please?
    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, free_to_fly!

    Use the subsitution $\displaystyle t=\sinh x$ to find:

    $\displaystyle \int_0^1 \sqrt {1 + t^2}\,dt$

    I used: .$\displaystyle t \,=\,\sinh x\quad\Rightarrow\quad dt \,=\,\cosh x\,dx$

    and that: .$\displaystyle 1 + \sinh^2\!x\:=\:\cosh^2\!x$

    So the integral becomes: .$\displaystyle \int_0^{\text{arcsinh 1}} \cosh^2\!x\,dx$

    and using the doule angle formula, I get: .$\displaystyle \frac{1}{2}\int_0^{\text{arcsinh 1}} (\cosh2x + 1)\,dx$

    Giving: . $\displaystyle \frac{1}{4}\sinh 2x + \frac{1}{2}x\,\bigg]^{\text{arcsinh 1}}_0$

    but then I can go no further because the answer had $\displaystyle \ln$ in it
    and I don't know where that came from.

    Evaluate and we get: .$\displaystyle \frac{1}{4}\,\sinh(2) + \frac{1}{2}\,\text{arcsinh 1} $

    . . and one of the basic identities is: .$\displaystyle \text{arcsinh }x \;=\;\ln\left|x + \sqrt{x^2+1}\right| $


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Where did that identity come from?
    . . You can derive it yourself.

    We have: .$\displaystyle \text{arcsinh x} \;=\:k$

    Then: .$\displaystyle \sinh k \:=\:x$

    From the definition: .$\displaystyle \frac{e^k - e^{-k}}{2} \:=\:x$

    Multiply by $\displaystyle 2e^k\!:\;\;e^{2k} - 1 \:=\:2xe^k\quad\Rightarrow\quad e^{2k} - 2xe^k - 1 \:=\:0$

    We have: .$\displaystyle (e^k)^2 - 2x(e^k) - 1 \:=\:0$ . . . a quadratic in $\displaystyle e^k$

    Quadratic Formula: .$\displaystyle e^k \;=\;\frac{-(-2x) \pm\sqrt{(-2x)^2 - 4(1)(-4)}}{2(1)} \;=\;x \pm\sqrt{x^2+1} $

    Since $\displaystyle e^k$ must be positive, we have: .$\displaystyle e^k \:=\:x + \sqrt{x^2+1}$

    Hence: .$\displaystyle k \;=\;\ln\left|x + \sqrt{x^2+1}\right| $

    Therefore: .$\displaystyle \text{arcsinh }x \;=\;\ln\left|x + \sqrt{x^2+1}\right| $ . . . ta-DAA!


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    Similarly, you can derive other identities:

    . . $\displaystyle \text{arccosh }x \;=\;\ln\left|x + \sqrt{x^2-1}\right|\quad \text{ for }x \geq 1$

    . . $\displaystyle \text{arctanh }x \;=\;\frac{1}{2}\ln\left|\frac{1+x}{1-x}\right|\quad \text{ for }|x| < 1$

    . . $\displaystyle \text{arcsech }x \;=\;\ln\left|\frac{1+\sqrt{1-x^2}}{x}\right|\quad \text{ for }0 < x \leq 1$


    Have I memorized these formulas? . . . certainly not!
    I have them on my handy-dandy Formula Sheets.
    [I need my brain cells to remember my name.]

    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member angel.white's Avatar
    Joined
    Oct 2007
    Posts
    724
    Awards
    1
    Soroban, I love you!

    I spent 3 hours last week figuing out how to differentiate secant cubed, but I never even considered the hyperbolic functions, our stupid book made such a huge fuss about the trigonometric that I lost sight of the bigger picture.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. integration by substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 3rd 2010, 11:12 PM
  2. Integration by Substitution Help
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 1st 2009, 08:28 PM
  3. integration by substitution xe^x^2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Oct 5th 2008, 10:48 PM
  4. integration by substitution
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 6th 2008, 05:22 PM
  5. integration by substitution
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Nov 19th 2006, 03:46 PM

Search Tags


/mathhelpforum @mathhelpforum