How do I integrate this in mathematica?

Questions
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1

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

2

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}]

plot.jpg
plot.jpg360×223 11.1 KB

Suvadip

3

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

4

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.

5

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

Untitled1.png
Untitled1.png764×412 9 KB

6

Hi!

No worries Suvadip.Thank you for your time and input.

And thank you Nasser. Interesting!

/Peter

7

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]*)