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Mathematica Calculus Wolfram Language

How do I integrate this in mathematica?

Peter Hemmälin

Posted 10 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 - 10sin x + sin^2 x)^2/(6 - 5sin x + sin^2 x) /Peter

POSTED BY: Peter Hemmälin

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Daniel Lichtblau

Daniel Lichtblau, Wolfram Research

Posted 10 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[2])])/Sqrt[2] - ( 128 ArcTan[(1 - 2 Tan[x/2])/Sqrt[3]])/Sqrt[3] + 15 Cos[x] - 1/4 Sin[2 x])

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[2])])/Sqrt[2] - (
 128 ArcTan[(1 - 2 Tan[x/2])/Sqrt[3]])/Sqrt[3] + 15 Cos[x] - 1/4 Sin[2 x]*)

POSTED BY: Daniel Lichtblau

Nasser M. Abbasi

Nasser M. Abbasi, student

Posted 10 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

POSTED BY: Nasser M. Abbasi

suvadip mandal

suvadip mandal, Indian Institute of Science Education and Research, Kolkata

Posted 10 years ago

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 BY: suvadip mandal

Peter Hemmälin

Posted 10 years ago

Hi! No worries Suvadip.Thank you for your time and input. And thank you Nasser. Interesting! /Peter

POSTED BY: Peter Hemmälin

Peter Hemmälin

Posted 10 years ago

Hi Suvadip I´m inclosing a document to show you how I have done. Please have a look and tell me if it´s right. /Peter Attachments:

POSTED BY: Peter Hemmälin

suvadip mandal

suvadip mandal, Indian Institute of Science Education and Research, Kolkata

Posted 10 years ago

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[1] = Integrate[(24 - 10 Sin[x] + (Sin[x])^2)/(6 - 5 Sin[x] + (Sin[x])^2),x] Out[1] = x + (3 ArcTan[(1 - 3 Tan[x/2])/(2 Sqrt[2])])/Sqrt[2] -16 ArcTan[(1 - 2 Tan[x/2])/Sqrt[3]])/Sqrt[3] 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[2])])/Sqrt[2] - (16 ArcTan[(1 - 2 Tan[x/2])/Sqrt[3]])/Sqrt[3], {x, -8, 8}] Suvadip

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[1] = Integrate[(24 - 10 Sin[x] + (Sin[x])^2)/(6 - 5 Sin[x] + (Sin[x])^2),x]
Out[1] = x + (3 ArcTan[(1 - 3 Tan[x/2])/(2 Sqrt[2])])/Sqrt[2] -16 ArcTan[(1 - 2 Tan[x/2])/Sqrt[3]])/Sqrt[3]

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[2])])/Sqrt[2] - (16 ArcTan[(1 - 2 Tan[x/2])/Sqrt[3]])/Sqrt[3], {x, -8, 8}]

enter image description here

Suvadip

POSTED BY: suvadip mandal

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