Integration using substitution

• Feb 12th 2008, 03:17 AM
free_to_fly
Integration using substitution
I'm stuck on the following question:

Use the subsitution t=sinhx to find:

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

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

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

Giving:
$
\
\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, 07:14 AM
Soroban
Hello, free_to_fly!

Quote:

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.]

• Feb 12th 2008, 07:28 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.