Integrate $\int(\log(\sin x \cos x))^n dx$ with hypergeometric function form

Question

Evaluate $$\int({\log(\sin x\cos x)})^{n} \, \mathrm{d}x$$ with result in hypergeometric function form

Could anyone help me with that?

Answer 1

Hint:

$\int(\log(\sin x\cos x))^n~dx=\int\left(\log\dfrac{\sin2x}{2}\right)^n~dx$

Let $u=\log\dfrac{\sin2x}{2}$ ,

Then $x=\dfrac{\sin^{-1}(2e^u)}{2}$

$dx=\dfrac{e^u}{\sqrt{1-4e^{2u}}}du$

$\therefore\int\left(\log\dfrac{\sin2x}{2}\right)^n~dx$

$=\int\dfrac{u^ne^u}{\sqrt{1-4e^{2u}}}du$

Case $1$: $|4e^{2u}|\leq1$

Then $\int\dfrac{u^ne^u}{\sqrt{1-4e^{2u}}}du$

$=\int\sum\limits_{m=0}^\infty\dfrac{(2m)!u^ne^{(2m+1)u}}{(m!)^2}du$

$=\int\sum\limits_{m=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(2m)!(2m+1)^ku^{n+k}}{(m!)^2k!}du$

Case $2$: $|4e^{2u}|\geq1$

Then $\int\dfrac{u^ne^u}{\sqrt{1-4e^{2u}}}du$

$=\int\dfrac{u^n}{\sqrt{e^{-2u}-4}}du$

$=\int\dfrac{u^n}{2i\sqrt{1-\dfrac{e^{-2u}}{4}}}du$

$=-\int\sum\limits_{m=0}^\infty\dfrac{i(2m)!u^ne^{-2mu}}{2^{4m+1}(m!)^2}du$

$=\int\sum\limits_{m=0}^\infty\sum\limits_{k=0}^\infty\dfrac{i(-1)^{k+1}(2m)!m^ku^{n+k}}{2^{4m-k+1}(m!)^2k!}du$