تعریف کنید:
$$f_n:=( \frac{sin^{2n-1}x+cos^{2n-1}x}{sin^{2n}x+cos^{2n}x} )^2$$
و انتگرال فوق را $I_n$ بنامید. حالا توجه کنید که:
$$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
بنابر این:
$$I_n= \int_\frac{-\pi}{4} ^ \frac{\pi}{4} f_n(x)dx= \int_\frac{-\pi}{4} ^ \frac{\pi}{4} f_n(-x)dx$$
$$ \Rightarrow 2I_n=\int_\frac{-\pi}{4} ^ \frac{\pi}{4}(f_n(x)+f(-x))dx=2\int_\frac{-\pi}{4} ^ \frac{\pi}{4} \frac{sin^{4n-2}x+cos^{4n-2}x}{(sin^{2n}x+cos^{2n}x)^2}dx$$
حالا با تغیر متغیر $z:=tanx$ داریم:
$$I_n=\int_{-1}^1 \frac{z^{4n-2}+1}{(1+z^{2n})^2}dz=2\int_0^1 \frac{z^{4n-2}+1}{(1+z^{2n})^2}dz$$
بار دیگر با تغییر متغیر $t:=\frac{1}{x} $ داریم:
$$\int_0^1 \frac{z^{4n-2}+1}{(1+z^{2n})^2}dz=\int_1^ \infty \frac{t^{4n-2}+1}{(1+t^{2n})^2}dt$$
بنابر این داریم:
$$2I_n=I_n+I_n=\int_0^1 \frac{z^{4n-2}+1}{(1+z^{2n})^2}dz+\int_1^ \infty \frac{t^{4n-2}+1}{(1+t^{2n})^2}dt$$
$$=\int_0^ \infty \frac{z^{4n-2}+1}{(1+z^{2n})^2}dz$$
$$=\int_0^ \infty \frac{z^{4n-2}}{(1+z^{2n})^2}dz+\int_0^ \infty \frac{1}{(1+z^{2n})^2}dz$$
حالا اگر تغییر متغیر $u:=z^{2n}$ را بکار ببریم داریم:
$$z=u^ \frac{1}{2n} \Rightarrow dz=\frac{1}{2n}u^{\frac{1}{2n}-1}$$
$$ \Rightarrow 2I_n= \frac{1}{2n} (\int_0^ \infty \frac{u^{(2-\frac{1}{2n})-1}}{(1+u)^{(2- \frac{1}{2n})+( \frac{1}{2n} )}}dz+\int_0^ \infty \frac{u^{(\frac{1}{2n})-1}}{(1+u)^{(\frac{1}{2n})+(2-\frac{1}{2n} )}}dz)$$
$$= \frac{1}{2n}(B(2- \frac{1}{2n}, \frac{1}{2n} )+B( \frac{1}{2n} ,2- \frac{1}{2n} )) = \frac{1}{n}B(2- \frac{1}{2n}, \frac{1}{2n})$$
$$= \frac{ \Gamma (2- \frac{1}{2n} ) \Gamma ( \frac{1}{2n} )}{ \Gamma (2- \frac{1}{2n} + \frac{1}{2n})}=\Gamma (2- \frac{1}{2n} ) \Gamma ( \frac{1}{2n} )$$
$$=(1- \frac{1}{2n} )\Gamma (1- \frac{1}{2n} ) \Gamma ( \frac{1}{2n} )=(1- \frac{1}{2n} ) \frac{\pi}{sin( \frac{\pi}{2n} )} $$
$$ \Rightarrow I_n= \frac{(2n-1)\pi}{2n^2sin( \frac{\pi}{2n} )}$$
$\Box$
