How do you evaluate the integral #int 1/xcos(lnx)dx#?

Question

5054 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-14T23:42:30

#int 1/x cos(lnx)dx = sin(lnx)+C#

Substitute:

#t=lnx#
#dt =dx/x#

so that:

#int 1/x cos(lnx)dx = int cost dt =sint +C#

Undo the substitution and we have:

#int 1/x cos(lnx)dx = sin(lnx)+C#

Answer 2 · 2017-02-14T23:43:27

The integral equals #sin(lnx) + C#

This is a substitution problem. Note that #d/dxlnx = 1/x#, so if we let #u = lnx#, then #du = 1/xdx# and #dx = xdu#.

#=>int 1/xcos(u) * xdu#

#=> int cosu du#

#=>sinu + C#

#=>sin(lnx) + C#

Hopefully this helps!