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 Foo Off-Topic chit chat with no general subject.

 04-20-09, 07:41 PM #1 phantomcow2 la vache fantôme Thread Starter     Join Date: Aug 2004 Location: NH Bikes: Posts: 6,266 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 1 Post(s) u substitution problem -- integration Find the indefinite integral for sec^3x*tanx using u substitution. I've done pretty well solving u substitution problems up to this point, but this problem continues to stump me; there is something I am missing. let u=tanx du = sec^2x*dx I figure that maybe I can "break apart" the sec^3x term into sec^2x*secx, giving me the following: sec^2x*secx*u*dx But du=sec^2dx so substituting gives secx*u*du But I can't find the anti derivative of this __________________ C://dos C://dos.run run.dos.run
 04-20-09, 11:06 PM #2 RubenX  Look! My Spine!     Join Date: Apr 2008 Location: Kissimmee, FL Bikes: Posts: 620 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 2 Post(s) There was a time when I knew this stuff. Sadly I've forgotten it all. My mind is tuned with boolean algebra and logic circuits ATM. Sorry...
 04-21-09, 06:44 AM #3 Tinuz Junior Member   Join Date: Oct 2008 Bikes: Posts: 21 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) assuming secx*u*du is correct, the integral to u would be 0.5*secx*u^2+C To see why, just reverse it. Again, not really a math wiz in the formal sense, so please check yourself. Also, this has been ages.
 04-21-09, 07:18 AM #4 Tinuz Junior Member   Join Date: Oct 2008 Bikes: Posts: 21 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) Maybe a better way is to write: sqrt(1+tan(x)^2)^3*tan(x) dx Because then if u=tan(x) you get: sqrt(1+u^2)^3*u du --> ((1+u^2)^0.5)^3*u du --> (1+u^2)^1.5*u du which eliminates the extra x, which I am not sure what to do about (integrate the result to x aftewards?). Finding the anitderivative should be easy from there on, but is has been ages, so you best do that yourself
 04-21-09, 07:38 AM #5 Tinuz Junior Member   Join Date: Oct 2008 Bikes: Posts: 21 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) Just a thought: (1+u^2)^1.5*u du looks like h(g(u))'*g(u)' with g(u)=1+u^2 thus h()'=()^1.5 making H()=(()^2.5)/2.5 ((1+u^2)^2.5)/2.5 However, the derivative of 1+u^2 is 2u so apparently there was a separate multiplicative term which ended up as 1/u. Then try to compensate for this.