Integration by parts of arcsec(x)

**Beeorz** · September 8th 2008, 06:32 PM

Having problems with the substitution. I don't see how it fits, although I feel like I am doing it correctly.

int arcsec(x) dx
=xarcsec(x) - int (x)/((x)(sqrt((x^2)-1)))
=xarcsec(x) - int 1/(sqrt(x^2)-1)

Here I try to use substitution but it doesn't seem to go well for me. Thanks in advance!

**mr fantastic** · September 8th 2008, 06:58 PM

Originally Posted by Beeorz

Having problems with the substitution. I don't see how it fits, although I feel like I am doing it correctly.

int arcsec(x) dx
=xarcsec(x) - int (x)/((x)(sqrt((x^2)-1)))
=xarcsec(x) - int 1/(sqrt(x^2)-1)

Here I try to use substitution but it doesn't seem to go well for me. Thanks in advance!

The easy way is to make a hyperbolic substitution.

The hard way is to make the substitution $x = \sec \theta$ .

Show the details of your calculation.

**Beeorz** · September 8th 2008, 07:14 PM

I've tried using:

u=arcsec(x)
du=1/(xsqrt((x^2)-1)))

Seems like the correct choice but I don't see how it would work. By hyperbolic sub-ing do you mean putting in:

arccosh(x) = 1/sqrt((x^2)-1)

**Soroban** · September 8th 2008, 07:14 PM

Hello, Beeorz!

Your work is correct . . .

$\int \text{arcsec}(x)\,dx$

. . $\begin{array}{cccccccc}u &=& \text{arcsec}(x) & & dv &=& dx \\ \\[-3mm]<br /> du &=&\dfrac{dx}{x\sqrt{x^2-1}} & & v &=& x \end{array}$

$x\!\cdot\!\text{arcsec}(x) - \int\frac{x}{x\sqrt{x^2-1}}\,dx \;=\;x\!\cdot\text{arcsec}(x) - \int \frac{dx}{\sqrt{x^2-1}}$ . . . . good work!

Let: . $x \:=\:\sec\theta \quad\Rightarrow\quad dx \:=\:\sec\theta\tan\theta\,d\theta$

$x\!\cdot\!\text{arcsec}(x) - \int\frac{\sec\theta\tan\theta\,d\theta}{\tan\thet a} \;=\;x\!\cdot\!\text{arcsec}(x) - \int\sec\theta\,d\theta \;=$ . $x\!\cdot\!\text{arcsec}(x) - \ln|\sec\theta + \tan\theta| + C$

Back-substitute: . $\sec\theta \:=\:x\quad\Rightarrow\quad \tan\theta \:=\:\sqrt{x^2-1}$

Therefore: . $\boxed{x\!\cdot\!\text{arcsec}(x) - \ln|x + \sqrt{x^2-1}| + C}$