How far should I trust Mathematica's Integrator?/ Why do I get two inconsistent answers?

Question

I am trying to solve the following integral: $$\int_0^{2\pi}\frac{\cos(2\theta)}{\sqrt{d^2+2\lambda^2-d^2\cos{2\theta}} (d^2+2\lambda^2+2\lambda^2\cos{2\theta})}\text{d}\theta$$. To do that I can find its primitive and evaluate the difference at the extremes:

$Assumptions = {θ ∈ Reals, d ∈ Reals, λ ∈ Reals};
T =  Cos[2θ]/(Sqrt[d^2 + 2 λ^2 -d^2 Cos[2θ]] (d^2 + 2 λ^2 + 2 λ^2 Cos[2θ]));
pr = Integrate[T, θ];
Int = pr/. θ-> 2π - pr/. θ-> 0

This gives as a result $0$ which means for any $d$,$\lambda$. However if I try a numerical integration for two given values of $d$,$\lambda$, say $d=\lambda=1$, which I should be able to compute as

NIntegrate[T /. d-> 1/.λ-> 1, {θ, 0, 2π}, AccuracyGoal -> Infinity]

I get quite a different result ($-0.477437$ in this case).

Can anyone tell me which result should I trust or what I am doing wrong?

---

Interestingly, when the same procedure above is used to compute other integrals the results are in agreement.

Answer 1

In the comments, the OP requested a closed-form solution. This approach appears to provide one. It uses the Rubi integration package for Mathematica (available at rulebasedintegration.org). Note that the Rubi integration command is "Int":

T = Cos[2 θ]/(Sqrt[d^2 + 2 λ^2 - d^2 Cos[2 θ]](d^2 + 2 λ^2 + 2 λ^2 Cos[2 θ]));
pr = Int[T, θ]

$\frac{\sqrt{\frac{d^2 (-\cos (2 \theta ))+d^2+2 \lambda ^2}{\lambda ^2}} F\left(\theta \left|-\frac{d^2}{\lambda ^2}\right.\right)}{2 \sqrt{2} \lambda ^2 \sqrt{d^2 (-\cos (2 \theta ))+d^2+2 \lambda ^2}}-\frac{\left(d^2+2 \lambda ^2\right) \sqrt{\frac{d^2 (-\cos (2 \theta ))+d^2+2 \lambda ^2}{\lambda ^2}} \Pi \left(\frac{4 \lambda ^2}{d^2+4 \lambda ^2};\theta \left|-\frac{d^2}{\lambda ^2}\right.\right)}{2 \sqrt{2} \lambda ^2 \left(d^2+4 \lambda ^2\right) \sqrt{d^2 (-\cos (2 \theta ))+d^2+2 \lambda ^2}}$

[$F$ is EllipticF, $\Pi$ is EllipticPi and $K$, seen in the next result, is EllipticK]

Rubi only calculates indefinite integrals. To obtain the definite integral, we substitute the limits of integration:

SymbInt = pr /. θ -> 2 π - pr /. θ -> 0

$\frac{\sqrt{2} K\left(-\frac{d^2}{\lambda ^2}\right)}{\left(\lambda ^2\right)^{3/2}}-\frac{\sqrt{2} \left(d^2+2 \lambda ^2\right) \Pi \left(\frac{4 \lambda ^2}{d^2+4 \lambda ^2}|-\frac{d^2}{\lambda ^2}\right)}{\left(\lambda ^2\right)^{3/2} \left(d^2+4 \lambda ^2\right)}$

Using JimB's nice code from the comment above for the numeric result from NIntegrate, and Plot3D for the symbolic result from Rubi (SymbInt), we can plot and compare the two for a range of values of d and λ. Here is the code used to plot the surfaces resulting from NIntegrate and Rubi, respectively:

surfaceN = Flatten[Table[{d, λ, NIntegrate[T, {θ, 0, 2 π}]}, {d, 1/10, 5, 1/10}, {λ, 1/10, 5, 1/10}], 1];
ListPlot3D[surfaceN, PlotRange -> {{1/10, 5}, {1/10, 5}}]
Plot3D[SymbInt, {d, 1/10, 5}, {λ, 1/10, 5}, PlotRange ->{{1/10, 5}, {1/10, 5}}]

The graphical results appear to be identical over this set of values:

One can also compare the results numerically:

TN = Table[NIntegrate[T, {θ, 0, 2 π}], {d,1/10, 5, 1/10}, {λ, 1/10, 5, 1/10}];
TS = Table[SymbInt, {d, 1/10, 5, 1/10}, {λ,1/10, 5, 1/10}] // N;
diff = TS - TN;
{Max@diff, Min@diff}

{3.0914*10^-7, -5.49857*10^-7}

Giving NIntegrate a higher WorkingPrecision brings its results closer to those of the symbolic form:

TN1 = Table[NIntegrate[T, {θ, 0, 2 π}, MinRecursion -> 2, WorkingPrecision -> 20], {d, 1/10, 5, 1/10}, {λ, 1/10, 5, 1/10}];
diff1 = TS - TN1;
{Max@diff1, Min@diff1}

{5.53158*10^-12, -7.72094*10^-12}

The OP will probably want to check the above approach for more general validity.