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# How do I integrate this in mathematica?

Posted 9 years ago
 Hi! How do I find an antiderivative to this expression? I´ve tried with the Integrate [...,x] but nothing comes out. Im supposed to answere with the coefficient in front of the x. Can´t seem to get it right. (24 - 10*sin x + sin^2 x)^2/(6 - 5*sin x + sin^2 x)  /Peter
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Posted 9 years ago
 I'd recommend trying with Mathematica format. ii = Integrate[(24 - 10 Sin[x] + Sin[x]^2)^2/(6 - 5 Sin[x] + Sin[x]^2), x] (*(135 x)/2 + (9 ArcTan[(1 - 3 Tan[x/2])/(2 Sqrt)])/Sqrt - ( 128 ArcTan[(1 - 2 Tan[x/2])/Sqrt])/Sqrt + 15 Cos[x] - 1/4 Sin[2 x]*) 
Posted 9 years ago
 Mathematica result contains discontinuities in the antiderivatives which is not desirable, as typically happen when using the common u=tan(x/2) substitution.Rubi package produces a smooth antiderivative. Here is the difference Hi Peter,I am really sorry. I did not notice the whole square thing in the numerator. I actually overlooked that. I have checked the notebook file you have attached and whatever you have done is correct. So I see you have got the answer.Suvadip.
Posted 9 years ago
 Hi!No worries Suvadip.Thank you for your time and input.And thank you Nasser. Interesting!/Peter
Posted 9 years ago
 Hi SuvadipI´m inclosing a document to show you how I have done. Please have a look and tell me if it´s right./Peter Attachments:
 Hi Peter,I have tried to evaluate the anti derivative i.e. integration of the function you have supplied.The result is the following. In = Integrate[(24 - 10 Sin[x] + (Sin[x])^2)/(6 - 5 Sin[x] + (Sin[x])^2),x] Out = x + (3 ArcTan[(1 - 3 Tan[x/2])/(2 Sqrt)])/Sqrt -16 ArcTan[(1 - 2 Tan[x/2])/Sqrt])/Sqrt It works fine. Perhaps you did not use the syntax properly.The plot what I have got is, Plot[x + (3 ArcTan[(1 - 3 Tan[x/2])/(2 Sqrt)])/Sqrt - (16 ArcTan[(1 - 2 Tan[x/2])/Sqrt])/Sqrt, {x, -8, 8}] Suvadip