how do i begin to solve this?
integral of (1/x^2)sec(1/x)tan91/x)dx
I tried substituting tan(1/x) for u but i couldn't get it to work...
You should know the derivatives of the six basic trig functions
recall that
if y = sec x
then dy/dx = secxtanx
I recognize that pattern in your integral and furthermore the derivative of 1/x, which is -1/x^2 sits out front as a factor
so it looks like we started with something like
y = sec (1/x)
then dy/dx = (-1/x^2)sec(1/x)tan(1/x)
the only difference between that and yours is the -1 in front, no big deal ....
integral of (1/x^2)sec(1/x)tan91/x)dx
= - sec(1/x) + c
I assumed that the last part of your expression contained a typo, the 9 should have been a '('