It really depends on the level at which you want to estimate this function. Do you want to end up with a nice closed expression? Do you simply need an expression to model the data? You can't be sure that an analytic solution exists. I tried the Rubi package (apmaths.uwo.ca/~arich) and it didn't give a solution. But you can always fit it to some curve.
{redata, imdata} =
Transpose[{{#1, Re[#2]}, {#1, Im[#2]}} & @@@
Table[
{parameter, NIntegrate[((Sin[u]^1.82 + (parameter^(-1))^0.63*Sin[u]^2.45)*
Sin[parameter + u]^1.82)/(parameter + Sin[u])^2, {u, 0,Pi}]}
, {parameter, 0.005, 1, .005}]];
ListLinePlot /@ {redata, imdata}

This suggests to me that we could model both the real and imaginary parts with a multi-exponential decay.
Grid[Table[
func = Sum[A[n] Exp[-B[n] x], {n, 1, nexp}];
params = Flatten[Table[{A[n], B[n]}, {n, nexp}]];
refit = NonlinearModelFit[redata, func, params, x];
imfit = NonlinearModelFit[imdata, func, params, x];
{Show[ListLinePlot[redata, ImageSize -> 300],
Plot[refit[x], {x, 0, 1}, PlotStyle -> {Dashed, Red}]],
ListPlot[Transpose[{redata[[All, 1]], refit["FitResiduals"]}],ImageSize -> 300],
Show[ListLinePlot[imdata, ImageSize -> 300],
Plot[imfit[x], {x, 0, 1}, PlotStyle -> {Dashed, Red}]],
ListPlot[Transpose[{imdata[[All, 1]], imfit["FitResiduals"]}],ImageSize -> 300]}
, {nexp, 2, 5}]]

So you could decide to ignore errors smaller than, say 0.5, which would allow you to ignore the imaginary part altogether, and take an answer with three exponential decay terms
$$51.6583 e^{-164.728 x}+15.9181 e^{-23.0781 x}+6.3396 e^{-2.9963 x}$$
A similar strategy would work for the upper integration limit.
parameter
not equal to zero, but you can't be sure that an analytic solution exists. I tried the Rubi package (apmaths.uwo.ca/~arich) and it didn't give a solution. $\endgroup$