suspenc3 Messages 400 Reaction score 0 Thread starter Aug 25, 2006 #1 Hi, I am kinda stuck on the following integral: [tex]\int\sqrt{\frac{1+x}{1-x}}dx[/tex] any hints?
TD Homework Helper Messages 1,016 Reaction score 0 Aug 25, 2006 #2 If you let [tex] y^2 = \frac{{1 + x}}{{1 - x}} \Leftrightarrow x = \frac{{y^2 - 1}}{{y^2 + 1}}[/tex] The integral will become fraction of rationals, losing the square root.
If you let [tex] y^2 = \frac{{1 + x}}{{1 - x}} \Leftrightarrow x = \frac{{y^2 - 1}}{{y^2 + 1}}[/tex] The integral will become fraction of rationals, losing the square root.
suspenc3 Messages 400 Reaction score 0 Aug 25, 2006 #4 or by making this substitution, the square root will be taken away
neutrino Messages 2,093 Reaction score 2 Aug 25, 2006 #5 Or you could do the trig substitution [tex]x = \sin\theta[/tex]
TD Homework Helper Messages 1,016 Reaction score 0 Aug 25, 2006 #6 suspenc3 said: so are you saying to substitute that for x? Yes, use that substitution to lose the square root. I already solved for x as well, which allows you to easily find dx in terms of dy by differentiating both sides.
suspenc3 said: so are you saying to substitute that for x? Yes, use that substitution to lose the square root. I already solved for x as well, which allows you to easily find dx in terms of dy by differentiating both sides.