# Integration using substitution

• Feb 12th 2008, 02:17 AM
free_to_fly
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?
• Feb 12th 2008, 06:14 AM
Soroban
Hello, free_to_fly!

Quote:

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

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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!

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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.]

• Feb 12th 2008, 06:28 AM
angel.white
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.