Question

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...

Answer 1

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 '('