Can someone point me in the right directon

  • Thread starter Thread starter gona
  • Start date Start date
  • Tags Tags
    Point
gona
Messages
9
Reaction score
0
the problem is to intigrate cot^{2}x

now i kow that cot^{2}x = \frac{cos^{2}x}{sin^{2}x}

now iv tryed various substitutions like usig the power reducton forumulas...
and then using double angle substiutions but nothin seems to work for me

so i was wondering if somone could just tell me where to start so i can
finish this question... been trying to figure this out for like a hour now

thanks in advance
 
Last edited:
Physics news on Phys.org
Use the identities 1+ cot2(x)= csc2(x) so that cot2(x)= csc2(x)- 1. Since you know that the derivative of cot(x) is csc2(x), that should be easy.
 
Wow that was suprisingly simple thank a lot yea i didnt know about that identity that was the first time iv used it. Thnx a lot great help
 
There are many pages that list trig identities. Whenever I work with trig functions I always just open a page and have a look at how the functions in the question are related. There are loads of pages, but I do like wikipedias as it has a list of the 6 trig functions in terms of the others which is very handy for some integration questions when trig substitution is needed.

http://en.wikipedia.org/wiki/Trigonometric_identity
 
Wow very nice I am going to bookmark this page right away thnx again!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top