Question

integral 1/(1-tan^2x)

Answer 1

\[\int\limits_{}^{}1/(1-\tan^2x)\]

Answer 2

I think we need to do u = 1-tan^2x I'm trying that now

Answer 3

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{1-\tan^2{x}}~dx}\)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{\dfrac{\cos^2x}{\cos^2x}-\dfrac{\sin^2x}{\cos^2{x}}}~dx}\)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{\dfrac{\cos^2x-\sin^2x}{\cos^2{x}}}~dx}\)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1\times \cos^2x}{\dfrac{\cos^2x-\sin^2x}{\cos^2{x}}\times \cos^2x}~dx}\)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{\cos^2x}{\cos^2x-\sin^2{x}}~dx}\)

Answer 4

no it is a mess my brain is not working right now

Answer 5

It's ok this problem is also making my brain hurt lol

Answer 6

Hmmm is it being the difference of two squares helpful at all?

Answer 7

How about cos2x = cos^2x - sin^2x
And cos^2x= cos2x+sin^2x

Answer 8

Empty i thought of that, but what would u then do

Answer 9

\[\int\limits_{}^{}(\cos2x + \sin^2x)/\cos2x\]

Answer 10

Integral of 1 + integral of sin^2x/cos2x

Answer 11

cos²x ?

Answer 12

or cos(2x) ?

Answer 13

I'm confused which cos you're talking about

Answer 14

both

Answer 15

cos2x = cos^2x - sin^2x
I used it to change cos^2x/(cos^2x - sin^2x)

Answer 16

I can't see what to do if we do use the difference of squares

Answer 17

that was a glitch it double posted idk why lol

Answer 18

yeah difference of squares doesn't get me anywhere, then what sub would I make?
and go into world's worse du=....

Answer 19

I used wolfram to solve it and based on the answer there should be a logarithm using u substitution

Answer 20

And a 1 somewhere

Answer 21

oh wait. If at the beginning, we do u = tanx
Then its du = sec^2xdx
du=(1+tan^2x)dx
So it would be
\[\int\limits_{}^{}1/((1-u^2)(1+u^2))\]

Answer 22

And that would be solvable with partial fractions I think

Answer 23

no master

Answer 24

sorry to dissapoint you, but your dx = (will be replaced by) cos²(idk what)

Answer 25

lol,

du=sec^2x dx
cos^2x du=dx
cos^2(tan\(^{-1}u\)) du=dx

Answer 26

u = tanx
du=sec^2xdx
cos^2xdu=dx

So \[\int\limits_{}^{}\cos^2x/(1-u^2)du\]

Answer 27

can not have that-:(
x's with u' and du like that

Answer 28

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

Answer 29

Even if you substitute it back? I didn't know

Answer 30

u= tan(x)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{\cos^2x}{1-u^2}~du}\)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1+\tan^2x)(1-u^2)}~du}\)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1+u^2)(1-u^2)}~du}\)

Answer 31

yes, sounds like it works.... no partial fractions is the intitive.

Answer 32

good job, i should say, master. indeed

Answer 33

now, partial fractions*

Answer 34

Thank you lol

Answer 35

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{1}{(1+\tan^2x)(1-u^2)}~du}\)

\(\large\color{slate}{\displaystyle\frac{1}{4}\int\limits_{~}^{~}\frac{2}{u^2+1}+\frac{1}{u+1}-\frac{1}{u-1}du}\)

Answer 36

oh, the tan^2x is u^2

Answer 37

so if that substitution actually works as we said, the rest should be easy.

Answer 38

Yup that matches what wolfram said the answer was.

Answer 39

Good job and thank you :)

Answer 40

Lol, I need to get back to one variable integration... this is a good exercise.

Answer 41

I had another idea, that I messaged to Empti

Answer 42

\(\large\color{black}{ \displaystyle \frac{1}{1-x} =\sum_{n=0}^{\infty}x^n }\)

\(\large\color{black}{ \displaystyle \frac{1}{1-\tan(x)} =\sum_{n=0}^{\infty}\tan^n(x) }\)

\(\large\color{black}{ \displaystyle \frac{1}{1-\tan^2(x)} =\sum_{n=0}^{\infty}\tan^{2n}(x) }\)

\(\large\color{black}{ \displaystyle \int\frac{1}{1-\tan^2(x)} dx=\int \sum_{n=0}^{\infty}\tan^{2n}(x)dx }\)

Answer 43

oh, no then, this becomes a difficult integral on the right

Answer 44

oh, you will learn it soon then.... it is just a power series representation thing very easy.

good luck

Answer 45

Thanks and to you as well

Answer 46

TH\(\color{blue}{\unicode{x13D49}}\)NK \(\color{blue}{\unicode{x3D4}}\)\(\color{blue}{O}\)U

Answer 47

cu