Solving Integral of 1/sqrt(1+x^2)

[karush]

$I=\int\frac{1}{\sqrt{1+x^2}}$
I followed this example but it surprised me
$x = \tan\theta \ \ dx = \sec^{2}\theta{d\theta}$
$$I=\int\left(\frac{1}{\sec\theta}\right){\sec^{2}\theta{d\theta}}
=\int{\sec\theta{d\theta}}$$

Why does this need to be done?
$$I=\int{\left(\frac{\sec^2\theta + \sec\theta\tan\theta}{\sec\theta + \tan\theta}\right){d\theta}}$$

Let, $(\sec\theta + \tan\theta) = u$
$(\sec^2\theta + \sec\theta\tan\theta)d\theta = du$

$$I=\int\frac{du}{u}$$

$$I=\ln{u}+c$$

$$I=\ln{\vert\sec\theta + \tan\theta\vert}+c$$

$$I=\ln{\vert\sqrt{1+\tan^2\theta} + \tan\theta\vert}+c$$

$I=\ln{\vert\sqrt{1+x^2} + x\vert}+c$, where $c$ is a constant

 

[Prove It]

$I=\int\frac{1}{\sqrt{1+x^2}}$
I followed this example but it surprised me
$x = \tan\theta \ \ dx = \sec^{2}\theta{d\theta}$
$$I=\int\left(\frac{1}{\sec\theta}\right){\sec^{2}\theta{d\theta}}
=\int{\sec\theta{d\theta}}$$

Why does this need to be done?
$$I=\int{\left(\frac{\sec^2\theta + \sec\theta\tan\theta}{\sec\theta + \tan\theta}\right){d\theta}}$$

Let, $(\sec\theta + \tan\theta) = u$
$(\sec^2\theta + \sec\theta\tan\theta)d\theta = du$

$$I=\int\frac{du}{u}$$

$$I=\ln{u}+c$$

$$I=\ln{\vert\sec\theta + \tan\theta\vert}+c$$

$$I=\ln{\vert\sqrt{1+\tan^2\theta} + \tan\theta\vert}+c$$

$I=\ln{\vert\sqrt{1+x^2} + x\vert}+c$, where $c$ is a constant

They did it because they recognised that $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}\theta}\,\left[ \sec{\left( \theta \right) } + \tan{ \left( \theta \right) } \right] = \sec{\left( \theta \right) } \tan{ \left( \theta \right) } + \tan^2{ \left( \theta \right) } \end{align*}$. As you say, this is not obvious. I personally would have written

$\displaystyle \begin{align*} \sec{ \left( \theta \right) } &= \frac{1}{\cos{\left( \theta \right) } } \\ &= \frac{ \cos{ \left( \theta \right) } }{\cos^2{ \left( \theta \right) } } \\ &= \frac{\cos{ \left( \theta \right) }}{1 - \sin^2{ \left( \theta \right) }} \end{align*}$

and then substitute $\displaystyle \begin{align*} u = \sin{ \left( \theta \right) } \implies \mathrm{d}u = \cos{ \left( \theta \right) } \,\mathrm{d}\theta \end{align*}$.On second thought, at the very start I would just substitute $\displaystyle \begin{align*} x =\sinh{(t)} \implies \mathrm{d}x = \cosh{(t)} \,\mathrm{d}t \end{align*}$...

$\displaystyle \begin{align*} \int{ \frac{1}{\sqrt{1 + x^2}}\,\mathrm{d}x} &= \int{ \frac{1}{\sqrt{1 + \sinh^2{(t)}}}\,\cosh{(t)}\,\mathrm{d}t } \\ &= \int{ \frac{\cosh{(t)}}{\sqrt{\cosh^2{(t)}}}\,\mathrm{d}t } \\ &= \int{ \frac{\cosh{(t)}}{\cosh{(t)}}\,\mathrm{d}t } \\ &= \int{ 1\,\mathrm{d}t } \\ &= t + C \\ &= \textrm{arsinh}\,\left( x \right) + C \end{align*}$

 

[karush]

Thanks that was very helpful..

Not sure why some of these examples
go off on weird trails

Your method makes more sense