Integral [(cosx)^3-(sinx)^2]dx/(cosx)2 Can i cancel the cosx leaving cosx in numerator before I integrate? If anyone could help with integrating this problem -
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Originally Posted by weezie23 Integral [(cosx)^3-(sinx)^2]dx/(cosx)2 Can i cancel the cosx leaving cosx in numerator before I integrate? If anyone could help with integrating this problem - You have the right idea. Split up the integrand: Now can you integrate ? hint: I hope this clarifies things! --Chris
integrate = sinx-tanx-1 b/c sec^2x -1 = tan^2x and Int.sec^2xdx= tanx+c Originally Posted by Chris L T521 You have the right idea. Split up the integrand: Now can you integrate ? hint: I hope this clarifies things! --Chris
Originally Posted by weezie23 integrate = sinx-tanx+1 b/c sec^2x -1 = tan^2x and Int.sec^2xdx= tanx+c VERY close...look at what I highlighted in red. First, it should be plus, not minus. Secondly, do you know what you forgot to do to the last term? Also, you forgot the constant of integration... --Chris
or i should say sinx-tanx + c
sinx-tanx -x +c
Originally Posted by weezie23 or i should say sinx-tanx + c Not quite... Fill in the blank... --Chris
Last edited by Chris L T521; July 30th 2008 at 06:10 PM. Reason: typo
sinx-tanx+x+c
Originally Posted by weezie23 sinx-tanx -x +c Make it plus, and then you have it --Chris
Originally Posted by weezie23 sinx-tanx+x+c Good. You have it now. --Chris
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