Question

What is the integral of (secx)/(9tan^2(x))

Answer 1

\[\large \int\limits \frac{\sec x}{9\tan^2x}dx\]Hmm there doesn't appear to be a nice `U substitution` so let's convert everything to sines and consines.

Answer 2

u-substitution doesnt work

Answer 3

\[\large \int\limits\limits \frac{\left(\dfrac{1}{\cos x}\right)}{9\left(\dfrac{\sin^2x}{\cos^2x}\right)}dx \qquad = \qquad \frac{1}{9}\int\limits\limits \left(\frac{1}{\cos x}\right)\left(\frac{\cos^2x}{\sin^2x}\right)dx\]

Answer 4

Understand how to proceed gal? :)
Do some cancellations with a couple cosines then you can do a nice easy `u sub` from there i think :D

Answer 5

simplify from zepdrix's last equation gives you \[1/9\int\limits_{}^{}(cotx * cscx)dx\]

Answer 6

which is simply -1/9 * cscx + c

Answer 7

Hmm that's what I got but Webassign says its wrong.

Answer 8

what's the website

Answer 9

Hmm well the answer is correct. Must just be a formatting error.
Are you typing otu the csc? Or using the button for it?

Answer 10

parenthesis aruond x.. write -1/9 * csc(x) + c