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\[\int\limits_{}^{}\frac{ 1 }{ \sec^2xtanx }dx\]

umm how did you get that first integral, sorry I'm confused

or would I just turn the sec^2x into (1-tan^2x)?

\[\int\limits_{}^{}cotxcos^2xdx\]

is that basically the same thing?

I hate calculus!!!

sorry just needed to vent

it just disappeared?

opppss. i forgot + c :)

from there I got this: \[47\int\limits_{}^{}\frac{ 1 }{ x^2\sqrt{x^2-(\frac{ 9 }{ 4 })^2} }\]

so then I did \[x=\frac{ 9 }{ 4 }\sec\]

that should be sec theta

somebody stop me if I'm screwing up lol

i have done it above @stottrupbailey
happy valentine's day , hahaha

And thus the current dilemma, the integral of sec^2tan

sorry, i didnt see the original problem.. i just see the first posting

from the equation :
x = 9/4 secθ or
secθ = 4x/9, reprsented in the figure :
|dw:1360828428409:dw|

thus, the result is
= 752/9 sinθ + c
= 752/9 * sqrt(16x^2 - 81)/(4x) + c