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Math Help - Integration using substitution

  1. #1
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    Integration using substitution

    I'm stuck on the following question:

    Use the subsitution t=sinhx to find:

    <br />
\<br />
\int_0^a {\sqrt {(1 + t} ^2 )dt} <br />
\<br />

    I used
    <br />
\<br />
\frac{{dt}}{{dx}} = \cosh x<br />
\<br />
and that
    <br />
\<br />
1 + \sinh ^2 x= \cosh ^2 x<br />
\<br />

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

    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?
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  2. #2
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    Hello, free_to_fly!

    Use the subsitution t=\sinh x to find:

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

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

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

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

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

    Giving: . \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 \ln in it
    and I don't know where that came from.

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

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


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


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

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

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

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

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

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

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

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

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

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


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


    Similarly, you can derive other identities:

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

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

    . . \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.]

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  3. #3
    Super Member angel.white's Avatar
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    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.
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