caseclosed Posted March 17, 2006 Share Posted March 17, 2006 I hope I am not bothering you guys with too many questions and post I am clueless, tried changing it to csc^2(x)csc(x) but doesn't help neither does csc(x)(1+cot^2(x)) Link to comment Share on other sites More sharing options...
Tom Mattson Posted March 18, 2006 Share Posted March 18, 2006 Try integrating by parts. Let [imath]u=\csc(x)[/imath] and let [imath]du=\csc^2(x)dx[/imath]. Link to comment Share on other sites More sharing options...
caseclosed Posted March 18, 2006 Author Share Posted March 18, 2006 Try integrating by parts. Let [imath]u=\csc(x)[/imath] and let [imath]du=\csc^2(x)dx[/imath']. I tried that and I did first time- U=csc(x) du=-csc(x)cot(x)dx dv=csc^2(x)dx v=-cot(x) second time - U=cot(x) du=-csc^2(x)dx dv=csc(x)cot(x)dx v=-csc(x) after doing all that it loops back to the problem so I get -csc(x)cot(x)+cot(x)csc(x)+I=I where I=integral of csc^3(x)dx so I subtract I, I get 0.... are the Us an Dv's suppose to stay consistent? Link to comment Share on other sites More sharing options...
Tom Mattson Posted March 18, 2006 Share Posted March 18, 2006 That is the correct method, but you should not have gotten 0. You must have made a small sign error. Link to comment Share on other sites More sharing options...
caseclosed Posted March 18, 2006 Author Share Posted March 18, 2006 here is my work, I don't see anywhere that I made the wrong sign. -csc(x)cot(x)-integral(-csc(x)(-cot^2(x)dx) -csc(x)cot(x)-[-cot(x)csc(x)-integral(-csc(x)(-csc^2(x))dx] -csc(x)cot(x)-cot(x)csc(x)+integral(csc^3(x)dx) so let integral(csc^3(x)dx)=I I get -csc(x)cot(x)-cot(x)csc(x)+I=I .... Link to comment Share on other sites More sharing options...
Tom Mattson Posted March 18, 2006 Share Posted March 18, 2006 here is my work' date=' I don't see anywhere that I made the wrong sign.-csc(x)cot(x)-integral(-csc(x)(-cot^2(x)dx) [/quote'] OK. -csc(x)cot(x)-[-cot(x)csc(x)-integral(-csc(x)(-csc^2(x))dx] Not sure how this came about. In your previous step you should have inserted [imath]\cot^2(x)=\csc^2(x)-1[/imath] to obtain the following. [math]\int\csc^3(x)dx=-\csc(x)\cot(x)-\int\csc^3(x)dx+\int\csc(x)dx[/math] It all works out straightforwardly from there. Link to comment Share on other sites More sharing options...
caseclosed Posted March 20, 2006 Author Share Posted March 20, 2006 awww, I did parts twice that is why did not work for me. Link to comment Share on other sites More sharing options...
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