how do i put ln[ sectheta + tantheta] in terms of x

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how do i put ln[ sectheta + tantheta] in terms of x

Mathematics
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a. There's no "x" in this expression. b. In order to put it in terms of a variable; it has to be in an equation form.
the ogrinal equation from solving it was intergal of dx/1-x^2 if that helps
true, isn't there a given x np?

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HaHaHA (breaking down due to extreme lough)
lol, what's so funny angoo?
I don't know
Ahhh, so you used a trig substitution. Now, go back to whatever value you used for x to solve the integral (in terms of theta), solve for sec(theta) and tan(theta) by drawing out the triangles, and put those values - in terms of x - in your final expression.
>_> seriously AG
just wrote that to make you smile
did you understand what Quantum said np65? ^_^
and it made me smile :)
so i have the triangle would that put me at 1-x^2+x then?
for the answer
what was your value of x when you integrated?
What trig substitution for x did you use to solve the integral?
i set x=sin
how did you set x = sin?
alright i have the intergal of dx1-x^2 i set u =sintheta and du = costheta dtheta i then subed in u and du
following that i got dtheta/costheta
oh I got it!\[x = \sin \theta\] and you know that sin x = opposite/hyp your opposite in this case = x hyp = 1 draw the triangle and you'll be able to continue from here ^_^
hence :\[\sec \theta= 1/\cos \theta\] \[\tan \theta = \sin \theta/ \cos \theta\] you have sin (theta), now find cos (theta) :) is it clear now?
so then the other side would be 1-x^2 or use the p-thag therom to get the last side making it the sqrroot(-x^2+1)
use the pyth.theorem :) that's it. you just don't have x, find it by applying the theorem , then find cos (theta)
so the last side will be : \[\sqrt{1- x^2}\] ^_^
awesome got it now
thanks
np :)

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