Solve the integral of 1/((secx)^2 + (tanx)^2)?

Question

Can anyone give me a hint of how to start it off in the right direction?

Answer 1

Start with tan^2 x + 1 = sec^2 x From trig identities. Then tan^2x + sec^2 x = 2 sec^2 -1.
Perhaps this is a "standard form" or some identity that is integrable.

Answer 2

well....first you can integrate around sums and differences, so now you have two integrals...(im gonna use the $ symbol for integal)
$ 1/(sec x)^2 and $ (tan x)^2

for the first one....
$ 1/(sec x)^2 is the same as $ (cos x)^2 so have fun integrating that (use properties of trig functions.hint: (cos x)^2= 1/2(1+cos (2x))

now just integrate (tan x)^2 which is the same as (sec x)^2-1 and you should know that the intergral of (sec x)^2 is tan x.....and when you integrate both, be sure the add your answers together....

I hope this helps....:)

Answer 3

exchange 6x^2 as t. Your indispensable sounds like ? 4xe^t dx On differentiating t=6x^2 you will get dt = 12x dx dx = dt / 12x exchange this and your indispensable sounds like ? a million/3 e^t dt The a million/3 is a persevering with take it out of the indispensable sign a million/3 ? e^t dt you recognize that ? e^x dx = e^x + c (the place c is an arbitrary consistent) a million/3 ? e^t dt = a million/3 e^t + c Re-substituting t a million/3 e^(6x^2) + c

Answer 4

uhhhh i shouldnt be looking at this. my head starts to hurt. hehe. well arent you supposed to start with the parentheses???

Answer 5

yeah.

pay attention in class.