M martin310015 Member Joined Dec 12, 2003 Messages 80 Gender Undisclosed HSC 2004 Jul 17, 2004 #1 how do u integrate cot^2(x) and 1/(-x^2+4x-3)^1/2...thanx
wogboy Terminator Joined Sep 2, 2002 Messages 653 Location Sydney Gender Male HSC 2002 Jul 17, 2004 #2 martin310015 said: how do u integrate cot^2(x) Click to expand... I cot^2(x) dx = I cosec^2(x) - 1 dx = I sec^2(x - pi/2) - 1 dx = tan(x - pi/2) - x + C = - cot(x) - x + C and 1/(-x^2+4x-3)^1/2...thanx Click to expand... Start off by completing the square in the denominator: I 1/sqrt(-x^2 + 4x - 3) dx = I 1/sqrt(-x^2 + 4x - 4 + 1) dx = I 1/(sqrt(1 - (x-2)^2) dx = arcsin(x-2) + C
martin310015 said: how do u integrate cot^2(x) Click to expand... I cot^2(x) dx = I cosec^2(x) - 1 dx = I sec^2(x - pi/2) - 1 dx = tan(x - pi/2) - x + C = - cot(x) - x + C and 1/(-x^2+4x-3)^1/2...thanx Click to expand... Start off by completing the square in the denominator: I 1/sqrt(-x^2 + 4x - 3) dx = I 1/sqrt(-x^2 + 4x - 4 + 1) dx = I 1/(sqrt(1 - (x-2)^2) dx = arcsin(x-2) + C