How Can I Find the Integral of Arcsec(x)dx Using IBP?

[pmg991818]

Can someone help me with finding the integral of arcsec (x) dx?

Thanks

Prakash

 

[chisigma]

Can someone help me with finding the integral of arcsec (x) dx?

Thanks

Prakash

Welcome on MHB pmg991818!...

... integrating by parts You obtain...

$\displaystyle \int \sec^{-1} x\ dx = x\ \sec^{-1} x - \int \frac {d x}{\sqrt{x^{2}-1}} = x\ \sec^{-1} x - \ln (\sqrt{x^{2}-1} + x) + c$

Kind regards

$\chi$ $\sigma$

 

[pmg991818]

Welcome on MHB pmg991818!...

... integrating by parts You obtain...

$\displaystyle \int \sec^{-1} x\ dx = x\ \sec^{-1} x - \int \frac {d x}{\sqrt{x^{2}-1}} = x\ \sec^{-1} x - \ln (\sqrt{x^{2}-1} + x) + c$

Kind regards

$\chi$ $\sigma$

Thanks chisigma

I will try and work it out. If you able to show it by steps would be appreciated.

Thank you again.

 

[chisigma]

Thanks chisigma

I will try and work it out. If you able to show it by steps would be appreciated.

Thank you again.

The essential point is that...

$\displaystyle \frac{d}{d x} \cosh^{-1} x = \frac{1}{\sqrt{x^{2}-1}}\ (1)$

... and that for x>1 is...

$\displaystyle \cosh^{-1} x = \ln (\sqrt{x^{2} - 1} + x)\ (2)$

Kind regards

$\chi$ $\sigma$

 

[Greg]

I used a substitution and IBP:

\(\displaystyle \int\sec^{-1}x\,dx\Leftrightarrow\sec u=x\Rightarrow u\sec u-\int\sec u\,du=u\sec u-\ln\left|\sec u+\tan u\right|+C\)

\(\displaystyle \int\sec^{-1}x\,dx=x\sec^{-1}x-\ln\left|x+\sqrt{x^2-1}\right|+C\)

 

[pmg991818]

I used a substitution and IBP:

\(\displaystyle \int\sec^{-1}x\,dx\Leftrightarrow\sec u=x\Rightarrow u\sec u-\int\sec u\,du=u\sec u-\ln\left|\sec u+\tan u\right|+C\)

\(\displaystyle \int\sec^{-1}x\,dx=x\sec^{-1}x-\ln\left|x+\sqrt{x^2-1}\right|+C\)

Thanks Rido.

 

[pmg991818]

I used a substitution and IBP:

\(\displaystyle \int\sec^{-1}x\,dx\Leftrightarrow\sec u=x\Rightarrow u\sec u-\int\sec u\,du=u\sec u-\ln\left|\sec u+\tan u\right|+C\)

\(\displaystyle \int\sec^{-1}x\,dx=x\sec^{-1}x-\ln\left|x+\sqrt{x^2-1}\right|+C\)

Thanks Greg 1313

 

[pmg991818]

Thanks Greg 1313

Greg,

I understand that you used IBP - arcsecx as u and dv =dx to get x. Then you have substituted for x sec u so, you get

uv = u sec u - \int sec u du. I presume you used this substitution on the basis of the right-angled triangle where sec u is x/1 giving us sec u = x. After integrating and getting ln absolute sec u + tan u, you have applied the trig substitution to get x = sec u and for tan you have applied \sqrt{x^2 -1} to get xarcsec{x} - ln|x +sqrt{x^2-1}| +C.

A bit easier to evaluate the indefinite integral than doing the whole thing by IBP.

Thank you.

Prakash

 

[Greg]

Then how do you evaluate

\(\displaystyle \int u\sec u\tan u\,du\)

without using IBP?

 

[pmg991818]

Then how do you evaluate

\(\displaystyle \int u\sec u\tan u\,du\)

without using IBP?

Not quite sure how the substitution part comes together. Greg, please clarify.

 

[Greg]

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

\(\displaystyle \sec u=x,\quad\sec u\tan u\,du=dx\)

So we have

\(\displaystyle \int u\sec u\tan u\,du\)

I think IBP is the most straightforward method to evaluate this integral.