Sunday, 31 March 2019

calculus - Integrate intp0ifrac3cosx+sqrt8+cos2xsinxxmathrmdx



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)+12 Li2(e2 x i),
where  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+arccot6y8y2.
So,
π03cosx+8+cos2xsinxx dx=06y(8+y2)(4+y2)(16+y2)(π2+arccot6y8y2)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+arccot6y8y2)dy=12i(2Li2(iy8+12)+Li2(iy6+13)+2Li2(iy6+23)+Li2(iy4+12)+Li2(2iy2i)Li2(2iy+2i)Li2(16i(y+2i))Li2(14i(y+2i))2(Li2(2iy4i)+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)(log5762log(y2+16))arctan(4y)+log(y2+4)arccot(6y8y2)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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