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anonymous
 one year ago
integral 1/(1tan^2x)
anonymous
 one year ago
integral 1/(1tan^2x)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}1/(1\tan^2x)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think we need to do u = 1tan^2x I'm trying that now

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{1\tan^2{x}}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{\dfrac{\cos^2x}{\cos^2x}\dfrac{\sin^2x}{\cos^2{x}}}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{\dfrac{\cos^2x\sin^2x}{\cos^2{x}}}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1\times \cos^2x}{\dfrac{\cos^2x\sin^2x}{\cos^2{x}}\times \cos^2x}~dx}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{\cos^2x}{\cos^2x\sin^2{x}}~dx}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1no it is a mess my brain is not working right now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It's ok this problem is also making my brain hurt lol

Empty
 one year ago
Best ResponseYou've already chosen the best response.1Hmmm is it being the difference of two squares helpful at all?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How about cos2x = cos^2x  sin^2x And cos^2x= cos2x+sin^2x

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Empty i thought of that, but what would u then do

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{}^{}(\cos2x + \sin^2x)/\cos2x\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Integral of 1 + integral of sin^2x/cos2x

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm confused which cos you're talking about

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0cos2x = cos^2x  sin^2x I used it to change cos^2x/(cos^2x  sin^2x)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I can't see what to do if we do use the difference of squares

Empty
 one year ago
Best ResponseYou've already chosen the best response.1that was a glitch it double posted idk why lol

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yeah difference of squares doesn't get me anywhere, then what sub would I make? and go into world's worse du=....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I used wolfram to solve it and based on the answer there should be a logarithm using u substitution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh wait. If at the beginning, we do u = tanx Then its du = sec^2xdx du=(1+tan^2x)dx So it would be \[\int\limits_{}^{}1/((1u^2)(1+u^2))\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0And that would be solvable with partial fractions I think

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1sorry to dissapoint you, but your dx = (will be replaced by) cos²(idk what)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1lol, du=sec^2x dx cos^2x du=dx cos^2(tan\(^{1}u\)) du=dx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0u = tanx du=sec^2xdx cos^2xdu=dx So \[\int\limits_{}^{}\cos^2x/(1u^2)du\]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1can not have that:( x's with u' and du like that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But 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
 one year ago
Best ResponseYou've already chosen the best response.0Even if you substitute it back? I didn't know

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1u= tan(x) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{\cos^2x}{1u^2}~du}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1+\tan^2x)(1u^2)}~du}\) \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1+u^2)(1u^2)}~du}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1yes, sounds like it works.... no partial fractions is the intitive.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1good job, i should say, master. indeed

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1now, partial fractions*

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1+\tan^2x)(1u^2)}~du}\) \(\large\color{slate}{\displaystyle\frac{1}{4}\int\limits_{~}^{~}\frac{2}{u^2+1}+\frac{1}{u+1}\frac{1}{u1}du}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1oh, the tan^2x is u^2

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1so if that substitution actually works as we said, the rest should be easy.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yup that matches what wolfram said the answer was.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Good job and thank you :)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Lol, I need to get back to one variable integration... this is a good exercise.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1I had another idea, that I messaged to Empti

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\large\color{black}{ \displaystyle \frac{1}{1x} =\sum_{n=0}^{\infty}x^n }\) \(\large\color{black}{ \displaystyle \frac{1}{1\tan(x)} =\sum_{n=0}^{\infty}\tan^n(x) }\) \(\large\color{black}{ \displaystyle \frac{1}{1\tan^2(x)} =\sum_{n=0}^{\infty}\tan^{2n}(x) }\) \(\large\color{black}{ \displaystyle \int\frac{1}{1\tan^2(x)} dx=\int \sum_{n=0}^{\infty}\tan^{2n}(x)dx }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1oh, no then, this becomes a difficult integral on the right

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1oh, you will learn it soon then.... it is just a power series representation thing very easy. good luck

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks and to you as well

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1TH\(\color{blue}{\unicode{x13D49}}\)NK \(\color{blue}{\unicode{x3D4}}\)\(\color{blue}{O}\)U
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