anonymous
  • anonymous
integral 1/(1-tan^2x)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{}^{}1/(1-\tan^2x)\]
anonymous
  • anonymous
I think we need to do u = 1-tan^2x I'm trying that now
SolomonZelman
  • SolomonZelman
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{1-\tan^2{x}}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{\dfrac{\cos^2x}{\cos^2x}-\dfrac{\sin^2x}{\cos^2{x}}}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{\dfrac{\cos^2x-\sin^2x}{\cos^2{x}}}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1\times \cos^2x}{\dfrac{\cos^2x-\sin^2x}{\cos^2{x}}\times \cos^2x}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{\cos^2x}{\cos^2x-\sin^2{x}}~dx}\)

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

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