Saturday, October 27, 2007

A 'Not so simple' Trig Problem

Now that we've cleared basic trig! Let's do Trig! With Integration! Yay!

Integral from 0 to Pi/4


(tan(x) - (tan(x)-1)/(tan(x)+1))dx

I would do it in math type, but I think that is simple enough.

Looking back at it (now that I know how to do it) it's not THAT hard, but like. How the hell was I supposed to know how to do it >.>


David Jin said...

Btw the answer is ...



Yep. How gay is that.

gee_cee0 said...

Lol... Ok firstly, sort of scared to try your questions now, not sure if what I know on my syllabus even covers everything I should know to solve this.

Anyway, this is how far I got... Please assume all intergrals from 0 to Pi/4, I won't sub the values to the end (which I haven't come to yet)

From int(tan(x) - (tan(x)-1)/(tan(x)+1))

= -ln(cos(x)) - int((tan(x)-1)/(tan(x)+1))

(Split the tan(x)'s into sin(x)/cos(x))

= -ln(cos(x)) - int((sin(x)-cos(x))/(sin(x)+cos(x))

(Used a shitload of identities here, including multiplying by conjugate, sin^2(x)+cos^2(x)=1, 2sin(x)cos(x) = sin(2x), sin^2(x)-cos^2(x)=-cos(2x))

= -ln(cos(x)) + int((1-sin(2x))/(cos2x))

Which then left me with a problem... I have no idea how to integrate sec(2x). At this point I gave up, perhaps you could help by telling me how I do that, or (more likely) where I went wrong?

P.S I think this is wrong, checking with online integrators gave me answers so horrific that I got scared out of studying and into playing games.

Don't bother with mathtype imo, I can understand calculator format ^^

David Jin said...

IB believe in using the most explicit and fastest ways of obtaining answers. Hence no splitting of Tan was used in this problem (Biggest hint ever)