Integrate
sqrt(1 +a^2*x^2) dx
Thanks
Try $\displaystyle x=\frac{tan(t)}{a}$
$\displaystyle dx=\frac{sec^2(t)}{a}dt$
$\displaystyle \int\sqrt{1+tan^2(t)}\frac{sec^2(t)}{a}dt$
$\displaystyle \frac{1}{a}*\int sec^3(t)dt$
$\displaystyle \frac{{sec(t)tan(t)}^3}{a}$
$\displaystyle t = tanh(ax)$
$\displaystyle \frac{{sec(tanh(ax))tan(tanh(ax))}^3}{a}$
$\displaystyle \frac{{sec(tanh(ax))(ax))}^3}{a}$
Hello, j-lee00!
Integrate: .$\displaystyle \int \sqrt{1 +a^2x^2}\,dx$
Let: $\displaystyle ax \:=\:\tan\theta\quad\Rightarrow\quad dx \:=\:\frac{1}{a}\sec^2\!\theta\,d\theta$
Then: .$\displaystyle \sqrt{1+a^2x^2} \:=\:\sec\theta$
Substitute: .$\displaystyle \int\sec\theta\left(\frac{1}{a}\sec^2\!\theta\,d\t heta\right) \;=\;\frac{1}{a}\int\sec^3\!\theta\,d\theta$
This integral requires Integration By Parts twice, or a standard formula.
. . $\displaystyle \frac{1}{2a}\bigg[\sec\theta\tan\theta + \ln\left|\sec\theta + \tan\theta\right|\bigg] + C$
Back-substitute . . .
$\displaystyle \text{Since }\tan\theta \:=\:\frac{ax}{1} \:=\:\frac{opp}{adj},\:\text{ then: }\:hyp \:=\:\sqrt{1+a^2x^2} \quad\Rightarrow\quad \sec\theta \:=\:\sqrt{1+a^2x^2}$
Answer: . $\displaystyle \frac{1}{2a}\bigg[ax\sqrt{1+a^2x^2} \:+ \:\ln\left|ax + \sqrt{1+a^2x^2}\right|\bigg] + C$
J Lee was trying to use a sinh substitution; I think the integration is easier with this substitution.
$\displaystyle x=\frac{\sinh{u}}{a}\ \Rightarrow\ \mathrm{d}x=\frac{\cosh{u}}{a}\,\mathrm{d}u$
$\displaystyle \begin{array}{rcl}
\therefore\ \int{\sqrt{1+a^2x^2}}\,\mathrm{d}x &=& \frac{1}{a}\int{\sqrt{1+\sinh^2{u}}\cdot\cosh{u}}\ ,\mathrm{d}x\\\\
{} &=& \frac{1}{a}\int{\cosh^2{u}}\,\mathrm{d}x\\\\
{} &=& \frac{1}{2a}\int{(\cosh{2u}+1)}\,\mathrm{d}x\\\\
{} &=& \frac{1}{2a}\,(\frac{1}{2}\sinh{2u}+u)+C\\\\
{} &=& \frac{1}{2a}\,(\sinh{u}\cosh{u}+u)+C\\\\
\end{array}$
Now you can substitute the x back (noting that $\displaystyle \sinh^{-1}{ax}=\ln{(ax+\sqrt{1+a^2x^2})}).$