Dave Posted March 19, 2003 Share Posted March 19, 2003 since i'm once again bored, I thought I'd share a problem with you. Solve :int: sqrt(1+sin(2x))dx I've seen a very nice way of doing this that made me kick myself after I'd seen it, so i'll let you all have a look at it before before spoiling the fun have fun. Link to comment Share on other sites More sharing options...
fafalone Posted March 19, 2003 Share Posted March 19, 2003 Just use :int:uv = uv - :int: v du u=1+sin(2x) v=sqrt(1+sin(2x)) du=2cos(2x) (chain rule) (1+sin(2x))(sqrt(1+sin(2x))) - :int: (sqrt(1+sin(2x)))(2cos(2x)) (sin(2x)+1)^(3/2) - :int: (cos(2x)/sqrt(sin(2x)+1) (sin(2x)+1)^(3/2) - sqrt(sin(2x)+1) which you can further simplify to (sin(2x))(sqrt(sin(2x)+1)) i'm sure i made a stupid mistake somewhere, but the general idea is right... and for some reason my calculator can't do this problem :/ Link to comment Share on other sites More sharing options...
fafalone Posted March 20, 2003 Share Posted March 20, 2003 Attempt number two: (sin(2x)+1)^(3/2) - :int: 2sqrt(sin(2x)+1)*cos(2x) (sin(2x)+1)^(3/2) - (2(sin(2x)+1))^(3/2))/3 === (sin(2x)+1)^(3/2))/3 There's still a stupid mistake, but this is closer. It usually takes me 5-6 attempts to solve a problem without stupid mistakes. Link to comment Share on other sites More sharing options...
Roark Posted March 20, 2003 Share Posted March 20, 2003 Dude, If you're bored then please take a look at my differential equation problem. I'm having trouble finding a good resourse on the net that lays out an introduction to differential equations and is full of worked problems. If you know of one, please share. Thank You! Link to comment Share on other sites More sharing options...
Dave Posted March 20, 2003 Author Share Posted March 20, 2003 nice method: you can use the fact that (sin(x))^2 + (cos(x))^2 = 1 to solve this problem. clearly this implies that: :int: sqrt((sin(x))^2 + sin(2x) + (cos(x))^2)dx = :int: sqrt((sin(x))^2 + 2sin(x)cos(x) + (cos(x))^2)dx but (sin(x))^2 + 2sin(x)cos(x) + (cos(x))^2 = (sin(x)+cos(x))^2, so the integral boils down to: :int: (sin(x) + cos(x))dx which i thought was rather nifty in regard to the differential equation problem, i have looked at it, but it is (as you say) a bit of a toughie. i may have a look at it in a bit. Link to comment Share on other sites More sharing options...
fafalone Posted March 20, 2003 Share Posted March 20, 2003 Trig identities are evil. Link to comment Share on other sites More sharing options...
fafalone Posted March 20, 2003 Share Posted March 20, 2003 sin(x) + cos(x) :neq: sqrt(1+sin(2x)) Link to comment Share on other sites More sharing options...
Dave Posted March 21, 2003 Author Share Posted March 21, 2003 if you want to be pedantic, then |sin(x) + cos(x)| = sqrt(1+sin(2x)) but even so, i thought it was a pretty method for solving a problem that looks fairly nasty. Link to comment Share on other sites More sharing options...
JoeDaWolf Posted May 7, 2003 Share Posted May 7, 2003 nice lil identity Link to comment Share on other sites More sharing options...
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