Please help me to solve this integral:
∫π03cosx+√8+cos2xsinxx dx.
I managed to calculate an indefinite integral of the left part:
∫cosxsinxx dx= xlog(2sinx)+12ℑ Li2(e2 x i),
where ℑ Li2(z) denotes the imaginary part of the dilogarithm. The corresponding definite integral ∫π0cosxsinxx dx diverges. So, it looks like in the original integral summands compensate each other's singularities to avoid divergence.
I tried a numerical integration and it looks plausible that
∫π03cosx+√8+cos2xsinxx dx?=πlog54,
but I have no idea how to prove it.
Answer
Let
y=3cosx+√8+cos2xsinx,
then, solving this with respect to x, we get
x=π2+arccot6y8−y2.
So,
∫π03cosx+√8+cos2xsinxx dx=∫∞06y(8+y2)(4+y2)(16+y2)(π2+arccot6y8−y2)dy.
The latter integral can be solved by Mathematica and yields πlog54.
Of course, we want to prove that the result returned by Mathematica is correct.
The following statement is provably true, that can be checked directly by taking derivatives of both sides:
∫6y(8+y2)(4+y2)(16+y2)(π2+arccot6y8−y2)dy=12i(2Li2(iy8+12)+Li2(iy6+13)+2Li2(iy6+23)+Li2(iy4+12)+Li2(2iy−2i)−Li2(−2iy+2i)−Li2(−16i(y+2i))−Li2(−14i(y+2i))−2(−Li2(−2iy−4i)+Li2(2iy+4i)+Li2(−18i(y+4i))+Li2(−16i(y+4i))))+π(12log(3(y2+4))+log(364(y2+16)))+log(4(y2+4))arctan(y4)−(log576−2log(y2+16))arctan(4y)+log(y2+4)arccot(6y8−y2)−arctan(2y)log12+arctan(y2)log2
The remaining part is to calculate lim and \lim\limits_{y\to\infty} of this expression, which I haven't done manually yet, but it looks like a doable task.
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