How far can I go with the integral $\int \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x, $ where $n\in N$?

In my answer, I have found the integral

$$\int \frac{d x}{1-\sin x \cos x} =\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C_0 $$ Next,

$$ \begin{aligned} & \int \frac{\sin x \cos x}{1-\sin x \cos x} d x \\ =& \int \frac{d x}{1-\sin x \cos x}-\int 1 d x \\ =& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+C_{1} \end{aligned} $$

and $$ \begin{aligned} & \int \frac{\sin ^{2} x \cos ^{2} x}{1-\sin x \cos x} d x \\ =& \int \frac{1-\left(1-\sin ^{2} x \cos ^{2} x\right)}{1-\sin x \cos x} d x \\ =& \int \frac{d x}{1-\sin x \cos x}-\int(1+\sin x \cos x) d x \\ =& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+\frac{\cos 2 x}{4}+C_2 \end{aligned} $$

Now I want to go further, $$ \begin{aligned} & \int \frac{\sin ^{3} x \cos ^{3} x}{1-\sin x \cos x} d x \\ =& \int \frac{1-\left(1-\sin ^{3} x \cos ^{3} x\right)}{1-\sin x \cos x} d x \\ =& \int \frac{d x}{1-\sin x \cos x}-\int\left(1+\sin x \cos x+\sin ^{2} x \cos ^{2} x\right) d x \\ =& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+\frac{\cos 2 x}{4}-\int \frac{\sin ^{2} 2 x}{4} d x \\ =& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+\frac{\cos 2 x}{4}-\frac{1}{4}\int\frac{1-\cos 4 x}{2} d x\\ =& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+\frac{\cos 2 x}{4}-\frac{1}{8}\left(x-\frac{\sin 4 x}{4}\right) +C\\=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-\frac{9}{8} x+\frac{\cos 2 x}{4}+\frac{\sin 4 x}{32} +C_3 \end{aligned} $$

Then I discovered that the integral

$$ I(n)=\int \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x $$ has a telescoping series

$$I(k+1)-I(k)=-\int \sin ^{k} x \cos ^{k} x d x$$

Hence $$ I(n)-I(1)=-\sum_{k=1}^{n-1} \frac{1}{2^{k}} \int\sin ^{k}(2 x) d x $$

We can conclude that

$$ I(n)=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x-\sum_{k=1}^{n-1} \frac{1}{2^{k}}\int\sin ^{k}(2 x) d x $$

Then I was stuck with the last sum.

By the way, I found that $$ \begin{array}{l} \displaystyle J(n):=\int_{0}^{\frac{\pi}{4}} \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x =\frac{2 \pi}{3 \sqrt{3}}-\frac{\pi}{4}-\sum_{k=1}^{n-1} \frac{1}{2^{k+1}} \int_{0}^{\frac{\pi}{2}} \sin ^{k} x d x \end{array} $$

Using the Wallis Formula, we can find the values of $J(n)$ for any natural number $n$. For example,

$$ \begin{array}{l} J(1)=\frac{2 \pi}{3 \sqrt{3}}-\frac{\pi}{4}=\frac{\pi}{36}(8 \sqrt{3}-9) \\ J(2)=\frac{\pi}{36}(8 \sqrt{3}-9)-\frac{1}{4} \\ J(3)=\frac{\pi}{36}(8 \sqrt{3}-9)-\frac{1}{4}-\frac{1}{8} \cdot \frac{\pi}{4}=\left(\frac{2}{3 \sqrt{3}}-\frac{9}{32}\right) \pi-\frac{1}{4}\\ \vdots \end{array} $$

My question is whether we can find a closed form for the last sum.

Solution 1:

Just for your curiosity.

As said in comments, the result is not very pretty.

Using $$I_k=\int_0^{\frac \pi 2} \sin^k(x)\,dx=\frac{\sqrt{\pi }}{2}\,\,\frac{\Gamma \left(\frac{k+1}{2}\right)}{\Gamma \left(\frac{k+2}{2}\right)}$$ $$S_n=\sum_{k=1}^{n-1} \frac{1}{2^{k+1}} \int_{0}^{\frac{\pi}{2}} \sin ^{k}( x)\, d x$$ $$S_n=\frac{8 \sqrt{3}-9}{36} \pi-\frac{\sqrt{\pi }}{ 2^{n+3}}\, T_n$$ $$T_n=2\frac{ \Gamma \left(\frac{n+1}{2}\right)}{\Gamma \left(\frac{n+2}{2}\right)}\, _2F_1\left(1,\frac{n+1}{2};\frac{n+2}{2};\frac{1}{4}\right)+\frac{ \Gamma \left(\frac{n+2}{2}\right)}{\Gamma \left(\frac{n+3}{2}\right)}\, _2F_1\left(1,\frac{n+2}{2};\frac{n+3}{2};\frac{1}{4}\right)$$ where appears the gaussian hypergeometric function.

However, the individual values of the $S_n$ are not bad. They write $$S_n=a_n+ \pi b_n$$ The $a_n$ form the sequence $$\left\{0,\frac{1}{4},\frac{1}{4},\frac{7}{24},\frac{7}{24},\frac{3}{10},\frac{3}{10}, \frac{169}{560},\frac{169}{560},\frac{1523}{5040},\frac{1523}{5040},\frac{133}{440},\cdots\right\}$$ and the $b_n$ form the sequence $$\left\{0,0,\frac{1}{32},\frac{1}{32},\frac{19}{512},\frac{19}{512},\frac{157}{4096}, \frac{157}{4096},\frac{5059}{131072},\frac{5059}{131072},\frac{40535}{1048576},\frac{40535}{1048576},\cdots\right\}$$

Edit

If you plan to integrate in the range $0\le x \le \frac \pi 2$ $$I_{k+1}-I_k=\frac{\sec ^{-(k+1)}(x) }{k+1}\,\, _2F_1\left(\frac{1-k}{2},\frac{k+1}{2};\frac{k+3}{2};\cos^2(x)\right)$$