อยากทราบวิธีการอินทิเกรตข้อนี้แบบสวยๆครับ
$\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 :happy: |
อ้างอิง:
$$f_a\left(x+\frac{\pi}{2}\right)=\dfrac{1}{\big(a^{\cos(x)}+1\big)\,\big(a^{-\sin(x)}+1\big)}=a^{\sin(x)}\,f_a(x)\,,$$ $$f_a\left(x+\pi\right)=\dfrac{1}{\big(a^{-\sin(x)}+1\big)\,\big(a^{-\cos(x)}+1\big)}=a^{\sin(x)}\,a^{\cos(x)}\,f_a(x)\,,$$ 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 $a>0$, $$\int_{\frac{5\pi}{4}}^{\frac{33\pi}{4}}\,f_a(x)\,\mathrm{d}x=\sum_{k=0}^{2}\,\int_{\frac{5\pi}{4}+2k\pi}^{\frac{5\pi}{4}+2(k+1 )\pi}\,f_a(x)\,\mathrm{d}x+\int_{8\pi-\frac{3\pi}{4}}^{8\pi+\frac{\pi}{4}}\,f_a(x)\,\mathrm{d}x=3\left(\frac{\pi}{2}\right)+\frac{\pi}{4}=\frac{7\pi}{4}\,.$$ |
เวลาที่แสดงทั้งหมด เป็นเวลาที่ประเทศไทย (GMT +7) ขณะนี้เป็นเวลา 18:15 |
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