# Integrate sqrt(1 +a^2*x^2)dx

• December 19th 2007, 04:04 AM
j-lee00
Integrate sqrt(1 +a^2*x^2)dx
Integrate

sqrt(1 +a^2*x^2) dx

Thanks
• December 19th 2007, 04:28 AM
colby2152
You want to do a trig substitution so that the square root term will disappear. Have you tried subbing tangent, secant, cotangent, or cosecant in for x?
• December 19th 2007, 04:32 AM
j-lee00
i have been trying x = a*sinh u
• December 19th 2007, 04:46 AM
colby2152
Quote:

Originally Posted by j-lee00
i have been trying x = a*sinh u

Try $x=\frac{tan(t)}{a}$

$dx=\frac{sec^2(t)}{a}dt$

$\int\sqrt{1+tan^2(t)}\frac{sec^2(t)}{a}dt$

$\frac{1}{a}*\int sec^3(t)dt$

$\frac{{sec(t)tan(t)}^3}{a}$

$t = tanh(ax)$

$\frac{{sec(tanh(ax))tan(tanh(ax))}^3}{a}$

$\frac{{sec(tanh(ax))(ax))}^3}{a}$
• December 19th 2007, 04:54 AM
JaneBennet
Quote:

Originally Posted by j-lee00
i have been trying x = a*sinh u

$x=\frac{\sinh{u}}{a}$ should work.
• December 19th 2007, 06:01 AM
Soroban
Hello, j-lee00!

Quote:

Integrate: . $\int \sqrt{1 +a^2x^2}\,dx$

Let: $ax \:=\:\tan\theta\quad\Rightarrow\quad dx \:=\:\frac{1}{a}\sec^2\!\theta\,d\theta$
Then: . $\sqrt{1+a^2x^2} \:=\:\sec\theta$

Substitute: . $\int\sec\theta\left(\frac{1}{a}\sec^2\!\theta\,d\t heta\right) \;=\;\frac{1}{a}\int\sec^3\!\theta\,d\theta$

This integral requires Integration By Parts twice, or a standard formula.

. . $\frac{1}{2a}\bigg[\sec\theta\tan\theta + \ln\left|\sec\theta + \tan\theta\right|\bigg] + C$

Back-substitute . . .

$\text{Since }\tan\theta \:=\:\frac{ax}{1} \:=\:\frac{opp}{adj},\:\text{ then: }\:hyp \:=\:\sqrt{1+a^2x^2} \quad\Rightarrow\quad \sec\theta \:=\:\sqrt{1+a^2x^2}$

Answer: . $\frac{1}{2a}\bigg[ax\sqrt{1+a^2x^2} \:+ \:\ln\left|ax + \sqrt{1+a^2x^2}\right|\bigg] + C$

• December 19th 2007, 06:15 AM
colby2152
J Lee,

I didn't work with the geometry, but Soroban basically completed what I started for you.

-Colby
• December 19th 2007, 07:00 AM
JaneBennet
J Lee was trying to use a sinh substitution; I think the integration is easier with this substitution.

$x=\frac{\sinh{u}}{a}\ \Rightarrow\ \mathrm{d}x=\frac{\cosh{u}}{a}\,\mathrm{d}u$

$\begin{array}{rcl}
\therefore\ \int{\sqrt{1+a^2x^2}}\,\mathrm{d}x &=& \frac{1}{a}\int{\sqrt{1+\sinh^2{u}}\cdot\cosh{u}}\ ,\mathrm{d}x\\\\
{} &=& \frac{1}{a}\int{\cosh^2{u}}\,\mathrm{d}x\\\\
{} &=& \frac{1}{2a}\int{(\cosh{2u}+1)}\,\mathrm{d}x\\\\
{} &=& \frac{1}{2a}\,(\frac{1}{2}\sinh{2u}+u)+C\\\\
{} &=& \frac{1}{2a}\,(\sinh{u}\cosh{u}+u)+C\\\\
\end{array}$

Now you can substitute the x back (noting that $\sinh^{-1}{ax}=\ln{(ax+\sqrt{1+a^2x^2})}).$
• December 19th 2007, 07:30 AM
j-lee00