# I integrate a real function, I want a real result [closed]

I want to count some integral in Mathematica (actually, check that this integral is zero...) I arrive to an answer which has a zero real part but has some imaginary part! Why? This could be a mistake but I suppose that I am not writting down the commands correctly... What do you think?

r[ϕ_] = 2*(1 - (1/3)^2)/(1 + (1/10) *Cos[ϕ])

Integrate[(Sqrt[(r'[ϕ])^2 + r[ϕ]^2])/(Sqrt[2 (-1 + 1/(r[ϕ]))]), {ϕ, 0, 2 π}]


Mathematica gives the following answer

0. - 13.028165432276161 I

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## closed as off-topic by Michael E2, m_goldberg, Yves Klett, Leonid Shifrin, RunnyKineMar 23 '14 at 18:15

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Have a look at Plot[2 (-1 + 1/(r[\[Phi]])), {\[Phi], 0, 2 \[Pi]}]. –  b.gatessucks Mar 23 '14 at 10:56
so it seems you are not integrating a real function... –  chris Mar 23 '14 at 11:08

Let's see in detail what the comments are talking about. Define:

r[x_] := 2*(1 - (1/3)^2)/(1 + (1/10)*Cos[x])


Compute function under your integral:

g[x_] = (Sqrt[(r'[x])^2 + r[x]^2])/(Sqrt[2 (-1 + 1/(r[x]))]) // FullSimplify


Plot Im and Re on domain of interest:

Plot[{Re@g[x], Im@g[x]}, {x, 0, 2 Pi}]


So you are indeed basically integrating the imaginary part of the function while its real part is zero.

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