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

of

(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 >.>

3 comments:

David Jin said...

Btw the answer is ...

...

Ln(2)

Yep. How gay is that.

Glen 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)