Question

Evaluate the Intergal

Answer 1

\[\int\limits_{}^{}\frac{ 1-\tan(\theta) }{ 1+\tan(\theta) }d \theta \]

Answer 2

I tried rationalizing but that does not help.

Answer 3

So I rationalized and simplified the numerator and denominator by completing the square and I got:

\[\int\limits\limits_{}^{}\frac{ (\tan(\theta)-1)^2 }{ (\tan(\theta)+1)^2-2 } d \theta\]

Answer 4

That just makes things harder though.

Answer 5

have you tried to change
\[\tan \theta = \frac{ \sin \theta }{ \cos \theta }\] ?

then simplify

Answer 6

Any reason for that?

Answer 7

Is it because it's derivatives are inside?

Answer 8

to me, you should time both by (1+ tan ). try it, I got the answer from that

Answer 9

I did that as well. Didn't work.

Answer 10

nope, numerator = sec^2, right?

Answer 11

One second.

Answer 12

denominator = (1+tan)^2, right?

Answer 13

Yes. I expanded it however.

Answer 14

and let u = tan you have du =sec^2 dtheta,

Answer 15

i think that's the correct method!
answer is ln | cos x +sin x | + c

Answer 16

Ohh shoot! I made an algebra error >., .

Answer 17

both those substitution makes it more complex! I tried them all!

Answer 18

@Loser66 , 1-tan^(x) is NOT sec^2(x) .

Answer 19

oops, sorry, let me check

Answer 20

However, this gives me an idea.

Answer 21

yes, I know what I m wrong, numerator = 1- tan^2 so break down to 1/ deno - tan^2 / deno
and then tan^2 = sec^2 -1 break down again = sec^2 /deno + 1/deno
get what i mean?

Answer 22

Once sec. Let em try something.

Answer 23

hahaha... try many times to solve. that's the way I do.

Answer 24

Not on an exam though :P .

Answer 25

so?? we practice here to quickly respond on exam, is it not right?

Answer 26

Good point.

Answer 27

when you see this stuff again, ok, wooooo! got it

Answer 28

@chihiroasleaf , I f convert everything to sine and cosine I goet: