Integration By Substitution AGAIN

Hi, im struggling on these qu's its not the concept, its the differentating if trig functions and things:

Integrate between 0 and infinite: 1\(x^2 + 4) dx where x = 2tanu

Integrate between 0 and 3: 1\Sqrt(9 - x^2) dx where x= 3sinu

Integrate betwen 1 and infinite: 1/(1 + x^2)^2/3 where x = tanu

Cheers also if you could tell me the derivates of secx etc and 2tanx 2secx ets that would be much appricated or have a resource (ie online forumlae book), that would be great.

Streety

Reply 1

Bezza

u=arctan(x/2), x = 2tanu, dx/du = 2(secu)^2, dx = 2(secu)^2 du and at x=0, u=0, x=infinity, u=pi/2
so int 1/(x^2 + 4) dx = int 1/[4(tanu)^2 + 4] * 2(secu)^2 du = int 1/2*(secu)^2/(secu)^2 since (tanu)^2 + 1 = (secu)^2
so we have int 1/2 du = u/2 evaluated between pi/2 and 0 = pi/4 (the indefinite integral would be 1/2*arctan(x/2) + c)

If you want to do differentiations or check the end results for your integrations, use http://www.calc101.com