integral 1/(1-tan^2x)

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integral 1/(1-tan^2x)

Mathematics
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\[\int\limits_{}^{}1/(1-\tan^2x)\]
I think we need to do u = 1-tan^2x I'm trying that now
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{1-\tan^2{x}}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{\dfrac{\cos^2x}{\cos^2x}-\dfrac{\sin^2x}{\cos^2{x}}}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{\dfrac{\cos^2x-\sin^2x}{\cos^2{x}}}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1\times \cos^2x}{\dfrac{\cos^2x-\sin^2x}{\cos^2{x}}\times \cos^2x}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{\cos^2x}{\cos^2x-\sin^2{x}}~dx}\)

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Other answers:

no it is a mess my brain is not working right now
It's ok this problem is also making my brain hurt lol
Hmmm is it being the difference of two squares helpful at all?
How about cos2x = cos^2x - sin^2x And cos^2x= cos2x+sin^2x
Empty i thought of that, but what would u then do
\[\int\limits_{}^{}(\cos2x + \sin^2x)/\cos2x\]
Integral of 1 + integral of sin^2x/cos2x
cos²x ?
or cos(2x) ?
I'm confused which cos you're talking about
both
cos2x = cos^2x - sin^2x I used it to change cos^2x/(cos^2x - sin^2x)
I can't see what to do if we do use the difference of squares
that was a glitch it double posted idk why lol
yeah difference of squares doesn't get me anywhere, then what sub would I make? and go into world's worse du=....
I used wolfram to solve it and based on the answer there should be a logarithm using u substitution
And a 1 somewhere
oh wait. If at the beginning, we do u = tanx Then its du = sec^2xdx du=(1+tan^2x)dx So it would be \[\int\limits_{}^{}1/((1-u^2)(1+u^2))\]
And that would be solvable with partial fractions I think
no master
sorry to dissapoint you, but your dx = (will be replaced by) cos²(idk what)
lol, du=sec^2x dx cos^2x du=dx cos^2(tan\(^{-1}u\)) du=dx
u = tanx du=sec^2xdx cos^2xdu=dx So \[\int\limits_{}^{}\cos^2x/(1-u^2)du\]
can not have that-:( x's with u' and du like that
But can't we just put the cos^2x as sec^2x in the denominator. And then sec^2x = 1+tan^2x = 1+u^2
Even if you substitute it back? I didn't know
u= tan(x) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{\cos^2x}{1-u^2}~du}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1+\tan^2x)(1-u^2)}~du}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1+u^2)(1-u^2)}~du}\)
yes, sounds like it works.... no partial fractions is the intitive.
good job, i should say, master. indeed
now, partial fractions*
Thank you lol
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1+\tan^2x)(1-u^2)}~du}\) \(\large\color{slate}{\displaystyle\frac{1}{4}\int\limits_{~}^{~}\frac{2}{u^2+1}+\frac{1}{u+1}-\frac{1}{u-1}du}\)
oh, the tan^2x is u^2
so if that substitution actually works as we said, the rest should be easy.
Yup that matches what wolfram said the answer was.
Good job and thank you :)
Lol, I need to get back to one variable integration... this is a good exercise.
I had another idea, that I messaged to Empti
\(\large\color{black}{ \displaystyle \frac{1}{1-x} =\sum_{n=0}^{\infty}x^n }\) \(\large\color{black}{ \displaystyle \frac{1}{1-\tan(x)} =\sum_{n=0}^{\infty}\tan^n(x) }\) \(\large\color{black}{ \displaystyle \frac{1}{1-\tan^2(x)} =\sum_{n=0}^{\infty}\tan^{2n}(x) }\) \(\large\color{black}{ \displaystyle \int\frac{1}{1-\tan^2(x)} dx=\int \sum_{n=0}^{\infty}\tan^{2n}(x)dx }\)
oh, no then, this becomes a difficult integral on the right
oh, you will learn it soon then.... it is just a power series representation thing very easy. good luck
Thanks and to you as well
TH\(\color{blue}{\unicode{x13D49}}\)NK \(\color{blue}{\unicode{x3D4}}\)\(\color{blue}{O}\)U
cu

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