RideTheLightnin
Member
Hey guys im just having a little confusion with integrating some trig
Ok Im in Fitzpatrick page 154, Exercise 26(d)
Take for example Q8
integral of dx/sqrt(2-x^2)
So I know its tan^-1, with the answer being tan^-1 (x/root2) + C, but since a is not on the top, just dx is, you have to put (1/root2) out the front. So the answer is
(1/root2)tan^-1 (x/root2) + C
But, then i look at Q11,
integral of dx/sqrt(1-9x^2)
Again it is tan^-1, you take the 9 out of the bottom and put 1/3 out the front, so you get 1/3.integral of dx/(1/9+x^2). So therefore a^2=1/9, a=1/3. So, because dx is on the top again (ie 1), and not a, you would have to put another 1/3 out the front woudlnt you? the book has the answer as 1/3.tan^-1(3x)+C, but wouldnt you have to take that third out since a is not on the top (like on the standard integral table?), so 1/9 is out the front?
I dont know if that made any sense, but any help would be greatly appreciated.
CHeers.
Ok Im in Fitzpatrick page 154, Exercise 26(d)
Take for example Q8
integral of dx/sqrt(2-x^2)
So I know its tan^-1, with the answer being tan^-1 (x/root2) + C, but since a is not on the top, just dx is, you have to put (1/root2) out the front. So the answer is
(1/root2)tan^-1 (x/root2) + C
But, then i look at Q11,
integral of dx/sqrt(1-9x^2)
Again it is tan^-1, you take the 9 out of the bottom and put 1/3 out the front, so you get 1/3.integral of dx/(1/9+x^2). So therefore a^2=1/9, a=1/3. So, because dx is on the top again (ie 1), and not a, you would have to put another 1/3 out the front woudlnt you? the book has the answer as 1/3.tan^-1(3x)+C, but wouldnt you have to take that third out since a is not on the top (like on the standard integral table?), so 1/9 is out the front?
I dont know if that made any sense, but any help would be greatly appreciated.
CHeers.
