Recent content by pmg991818

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

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...

Thanks Greg 1313

Thanks Rido.

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

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