in (sin(x) cos(x)) dx let u=cos(x) du=-sin(x) dx answer is -1/2cos^2(x) (a) if let u=sin(x) du=cos(x) dx then answer is 1/2*sin^2(x) (b) if we put it into an right angle triangle length 3,4,5 then (a) get -0.18 but (b) get 0.32
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Originally Posted by oponion in (sin(x) cos(x)) dx let u=cos(x) du=-sin(x) dx answer is -1/2cos^2(x) (a) if let u=sin(x) du=cos(x) dx then answer is 1/2*sin^2(x) (b) if we put it into an right angle triangle length 3,4,5 then (a) get -0.18 but (b) get 0.32 You have forgotten about the arbitrary constant that is added to each integral. Both answers are correct up to a different constant, since .
thank you very much because my answer is different from the solution and i use x=1,2 sub into my answer and the solution and get different answer now i understand thank you
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