อยากทราบวิธีการอินทิเกรตข้อนี้แบบสวยๆครับ

[kompaui]

$\int_{\frac{5\pi}{4}}^{\frac{33\pi}{4}}\,\frac{1}{(2^{sin{x}}+1)(2^{cos{x}}+1)}dx $
ผมลองแทนที่ $y=\frac{19\pi }{4} -x$ ครับ แต่ไม่รู้มาถูกทางหรือป่าว

[Anton]

Problem. Evaluate the definite integral
$$\int_{\frac{5\pi}{4}}^{\frac{33\pi}{4}}\,\frac{1}{\left(2^{\sin(x)}+1\right)\,\left(2^{\cos(x)}+1\right)}\,\mathrm{d}x\,.$$

Fix $a>0$. Let $f_a(x):=\dfrac{1}{\big(a^{\sin(x)}+1\big)\,\big(a^{\cos(x)}+1\big)}$ for all $x\in\mathbb{R}$. Observe that
$$f_a\left(x+\frac{\pi}{2}\right)=\dfrac{1}{\big(a^{\cos(x)}+1\big)\,\big(a^{-\sin(x)}+1\big)}=a^{\sin(x)}\,f_a(x)\,,$$
$$f_a\left(x+\pi\right)=\dfrac{1}{\big(a^{-\sin(x)}+1\big)\,\big(a^{-\cos(x)}+1\big)}=a^{\sin(x)}\,a^{\cos(x)}\,f_a(x)\,,$$
and
$$f_a\left(x+\dfrac{3\pi}{2}\right)=\dfrac{1}{\big(a^{-\cos(x)}+1\big)\,\big(a^{\sin(x)}+1\big)}=a^{\cos(x)}\,f_a(x)\,.$$
Thus,
$$f_a(x)+f_a\left(x+\frac{\pi}{2}\right)+f_a(x+\pi)+f_a\left(x+\dfrac{3\pi}{2}\right)=\left(1+a^{\sin(x)}+a^{\sin(x)}\,a^{\cos(x )}+a^{\cos(x)}\right)\,f_a(x)=1\,.$$
Therefore, for any $\theta\in\mathbb{R}$,
$$\int_{\theta}^{2\pi+\theta}\,f(x)\,\mathrm{d}x=\int_{\theta}^{\theta+\frac{\pi}{2}}\,\Biggl(f_a(x)+{f_a \left(x+\frac{\pi}{2} \right)}+f_a(x+\pi)+{f_a \left(x+\dfrac{3\pi}{2} \right)} \Biggr)\,\mathrm{d}x=\frac{\pi}{2}\,.$$
Furthermore, observe that
$$f_a(x)=f_a\left(\dfrac{\pi}{2}-x\right)\,.$$
Hence,
$$\int_{2n\pi-\frac{3\pi}{4}}^{2n\pi+\frac{\pi}{4}}\,f_a(x)\,\mathrm{d}x=\int_{-\frac{3\pi}{4}}^{\frac{\pi}{4}}\,f_a(x)\,\mathrm{d}x=\frac{1}{2}\,\int_{-\frac{3\pi}{4}}^{\frac{5\pi}{4}}\,f_a(x)\,\mathrm{d}x=\frac{1}{2}\left(\frac{\pi}{2}\right)=\frac{\pi}{4}$$
for all integers $n$. Consequently, for all $a>0$,
$$\int_{\frac{5\pi}{4}}^{\frac{33\pi}{4}}\,f_a(x)\,\mathrm{d}x=\sum_{k=0}^{2}\,\int_{\frac{5\pi}{4}+2k\pi}^{\frac{5\pi}{4}+2(k+1 )\pi}\,f_a(x)\,\mathrm{d}x+\int_{8\pi-\frac{3\pi}{4}}^{8\pi+\frac{\pi}{4}}\,f_a(x)\,\mathrm{d}x=3\left(\frac{\pi}{2}\right)+\frac{\pi}{4}=\frac{7\pi}{4}\,.$$