# Thread: Integral (secx)^-1

1. ## Integral (secx)^-1

$\displaystyle /int sec^{-1}xdx$

How do I start?

2. Originally Posted by circuscircus
$\displaystyle /int sec^{-1}xdx$

How do I start?
Start by expressing the question clearly. Presumably you want:

integral arcsec(x) dx.

Start by putting u=arcsec(x), then sec(u)=x, or 1/cos(u)=x.

RonL

3. $\displaystyle u=arcsecx; sec u = x$

$\displaystyle \int arcsecx dx$

$\displaystyle \int sec u = \int x$

$\displaystyle ln|sec u + tan u| + C_{1} = 1/2 x^2 + C_{2}$

$\displaystyle ln|sec(arcsecx) + tan(arcsecx)| = 1/2 x^2 + C_{3}; c_{3}=C_{2}-C_{1}$\

$\displaystyle ln|x + \sqrt{x^2-1}| = 1/2 x^2 + C_{3}; C=-C_{3}$

$\displaystyle = ln|x + \sqrt{x^2-1}| - 1/2 x^2 + C$

correct?

4. Should be:

$\displaystyle xsec^{-1}(x)-ln|x+\sqrt{x^{2}-1}|+C$

or some form thereof.

5. Originally Posted by circuscircus
$\displaystyle u=arcsecx; sec u = x$

$\displaystyle \int arcsecx dx$

$\displaystyle \int sec u = \int x$

$\displaystyle ln|sec u + tan u| + C_{1} = 1/2 x^2 + C_{2}$

$\displaystyle ln|sec(arcsecx) + tan(arcsecx)| = 1/2 x^2 + C_{3}; c_{3}=C_{2}-C_{1}$\

$\displaystyle ln|x + \sqrt{x^2-1}| = 1/2 x^2 + C_{3}; C=-C_{3}$

$\displaystyle = ln|x + \sqrt{x^2-1}| - 1/2 x^2 + C$

correct?
No, it should be,

$\displaystyle xarcsecx-ln|x + \sqrt{x^2-1}|+C$

6. Where does xarcsecx come from? Also what happens to 1/2 x^2?

7. circus, you need to apply integration by parts

8. Hello, circuscircus!

polymerase is absolutely correct . . .

$\displaystyle \int \sec^{\text{-}1}\!(x)\,dx$

Integrate by parts . . .

$\displaystyle \begin{array}{ccccccc}u & = & \sec^{\text{-}1}(x) & \quad & dv & = & dx \\ du & = & \frac{dx}{x\sqrt{x^2-1}} & \quad & v & = & x \end{array}$

Then we have: .$\displaystyle x\!\cdot\!\sec^{\text{-}1}(x) - \int\frac{dx}{\sqrt{x^2-1}}$

Trig Substitution: .$\displaystyle x\,=\,\sec\theta\quad\Rightarrow\quad dx \,=\,\sec\theta\tan\theta\,d\theta,\;\;\sqrt{x^2-1}\,=\,\tan\theta$

. . Substitute: .$\displaystyle x\!\cdot\!\sec^{\text{-}1}(x) - \int\frac{\sec\theta\tan\theta\,d\theta}{\tan\thet a} \;=\;x\!\cdot\!\sec^{\text{-}1}(x) - \int\sec\theta\,d\theta$

. . and we have: .$\displaystyle x\!\cdot\!\sec^{\text{-}1}(x) - \ln|\sec\theta + \tan\theta| + C$

Back-substitute: .$\displaystyle x\!\cdot\!\sec^{\text{-}1}(x) - \ln|x + \sqrt{x^2-1}| + C$

9. Originally Posted by Soroban
$\displaystyle \int\frac{dx}{\sqrt{x^2-1}}$
Here's another attempt

Set $\displaystyle u=x+\sqrt{x^2-1}\implies du=1+\frac x{\sqrt{x^2-1}}\,dx\,\therefore\,du=\frac u{\sqrt{x^2-1}}\,dx,$ the integral becomes to

$\displaystyle \int\frac1{\sqrt{x^2-1}}\,dx=\int\frac1u\,du=\ln|u|+k$

Back substitute and we are done.