Question

Why is this integral of tan incorrect?

Note: change all x on my 3rd and 4th comment apart from those on the very first line of the 3rd comment to z.

Answer 1

\[\int \frac{\sin(x)}{\cos(x)}dx=\int \frac{\sin(x) \cos(x)}{\cos^2(x)}dx=\int \frac{\sin(x) \cos(x)}{1-\sin^2(x)}dx\]
\[z= \sin(x), dz=\cos(x)dx\]

\[=\int \frac{z }{1-z^2}dz\]
\[=\frac{-1}{2}\ln(-\cos(x))+k\]

Answer 2

I'll justify the end step:

Answer 3

\[\int \frac{x}{x+b} dx=-b \ln(x+b)+x\]
\[-\int \frac{z}{z^2-1} dx=- \frac{1}{2} \left(\int \frac{zdz}{z-1} -\int \frac{zdz}{z+1} \right) dx\]
\[=- \frac{1}{2} \left(-(-1) \ln(x+(-1))+x-(-(1) \ln(x+1)+x) \right) dx\]
\[=- \frac{1}{2} \left(\ln(x-1))+x+\ln(x+1)-x \right) dx\]
\[=- \frac{1}{2} \left(\ln(x-1))+\ln(x+1) \right) dx\]
\[=- \frac{1}{2} \left(\ln(x-1)(x+1) )\right) dx\]
\[=- \frac{1}{2} \left(\ln(x^2-1)\right) dx\]

Answer 4

\[=-\frac{1}{2}\ln(\sin^2(x)-1)=-\frac{1}{2}\ln(-\cos(x))\]

Answer 5

There are 2 minus signs that really shouldn't be there. Where did I make the mistake?

Answer 6

there must be cos^2x

Answer 7

Of course.

Answer 8

let me explain

Answer 9

By the way, I know the conventional way of integrating tan.

Answer 10

your first substitution is sinx, second is 1-z^2

Answer 11

Really? I think I just substituted once.

Answer 12

Sorry, I switched notation from \[z\] to \[x\] midway through. That was unintentional.

Answer 13

no there two substitutions

Answer 14

wait i'll write full form

Answer 15

Read everything on my third comment (except the very first line) as z.

Answer 16

\[\int\limits_{}\frac{ z }{ 1-z ^{2} }dz\]

Answer 17

u=1-z^2 du=-2z

Answer 18

\[-\frac{ 1 }{ 2 }\int\limits_{}\frac{ 1 }{ u }du\]

Answer 19

You misunderstand me, I'm perfectly aware how to do it that way: I'm more interested in seeing where my mistake was.

Answer 20

that equals \[-\frac{ 1 }{ 2 }\ln \left| u \right|=-\frac{ 1 }{ 2 }\ln \left| 1-z ^{2} \right|\]

Answer 21

z=sin^2

Answer 22

1-sin^2=cos^2

Answer 23

got it?

Answer 24

I already got it. You misunderstood me, I'm perfectly aware how to do it that way: I'm more interested in seeing where my mistake was.

Answer 25

your mistake was in partial fractions

Answer 26

it was incorrect

Answer 27

\[\frac{2}{(z-1)(z+1)}=-\frac{(z-1)}{(z-1)(z+1)}+\frac{(z+1)}{(z-1)(z+1)}=-\frac{1}{z+1}+\frac{1}{z-1}\]

Answer 28

That part?

Answer 29

the first part, when you just started to split it into to equations. It seems like that part (that you showed), but im sure there is a discrepancy

Answer 30

there was z instead of 2

Answer 31

No, the z was on both sides

Answer 32

I'm pretty sure that part was correct, it's probably because I forgot to take the absolute vale of cos^2 inside ln.

Answer 33

I'm still not sure why you have to take absolute values of the thing inside log

Answer 34

What's wrong with it being complex-valued?

Answer 35

yeah, seems your answer is also correct, don't forget that different method of solving integration may result in different answers

Answer 36

sorry, i was confused before