It's a problem that I found interesting because it gives an approximation of $\frac{\pi}{2}$ and a sequence of term .I can add to my problem a bit of elementary number theory furthermore.Finally and maybe we can make a link with the irrationality of $\pi$
We have that :
$$\int_{0}^{\frac{\pi}{2}}\frac{(\sin(x)-1)^n}{(\sin(x)+1)^n}dx=\frac{\pi}{2}-\frac{a}{b}$$ Where $n\geq 2$ a natural number and $a,b$ natural number to find.
Remark (number theory)
1)$b$ is always dividing by $5$
2)$a$ is always a composite number .
Remark (irrationality)
When $n$ increasing the integrals tends to $0$ so the fraction tends to $\frac{\pi}{2}$.Unfortunatly $a,b$ are choose arbitrary so it's inconvenient if we want to prove the irrationality of $\pi$ and caluclate the integral is delicate.
Remark sequence
My questions :
Can someone prove these remarks ?
Is the sequence referenced and is there a recurrence formula?
Can we prove the irrationality of $\pi$ with the integral ?
Thanks in advance !
Last remark :
$$\int_{0}^{\frac{\pi}{2}}\frac{(\sin(x)-1)^{n-1}}{(\sin(x)+1)^n}dx=\pm\frac{1}{2n-1}$$
So integration by parts is your friends in this case .
$$\begin{align} \int_0^{\pi/2}\left(\frac{\sin x-1}{\sin x+1}\right)^n dx &=\int_0^{\pi/2}\left(\frac{\cos x-1}{\cos x+1}\right)^n dx =\int_0^{\pi/2}\left(-\tan^2\frac x2\right)^n dx\\ &=2\int_0^{1}\frac{(-t^2)^n}{1+t^2} dt =2\int_0^{1}\left(\frac1{1+t^2}-\frac{1-(-t^2)^n}{1+t^2}\right) dt\\ &=\frac{\pi}2-2\int_0^1 \sum_{k=0}^{n-1}(-1)^kt^{2k} dt =\frac{\pi}2-2\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}. \end{align} $$ Thus, $$\frac ab=2\sum_{k=0}^{n-1}\frac{(-1)^k}{2k+1}.$$
The fact that the limit of the above sum for $n\to\infty$ is equal to $\frac\pi4$ is very well-known and as far as I know never was used for a proof of the irrationality of $\pi$.