อ้างอิง:
ข้อความเดิมเขียนโดยคุณ nooonuii
35. ตอบ $\dfrac{7\pi}{4}$ ครับ
เปลี่ยนตัวแปรโดยให้ $u=x-\dfrac{25\pi}{4}$ จะได้
$\displaystyle{\int_{\frac{25\pi}{4}}^{\frac{53\pi}{4}}\dfrac{1}{(1+2^{\cos{x}})(1+2^{\sin{x}})}\,dx=\int_0^{7\pi}\dfrac{1}{(1+2 ^{\cos{(x+\frac{\pi}{4})}})(1+2^{\sin{(x+\frac{\pi}{4})}})}\,dx}$
ให้
$f(x)=\dfrac{1}{(1+2^{\cos{(x+\frac{\pi}{4})}})(1+2^{\sin{(x+\frac{\pi}{4})}})}$
$g(x)=\dfrac{1}{(1+2^{\cos{x}})(1+2^{\sin{x}})}$
จะได้ว่า
$\displaystyle{\int_0^{\frac{\pi}{4}}f(x)\,dx=\int_0^{\frac{\pi}{4}}f(x)\,dx}$
$\displaystyle{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}f(x)\,dx=\int_0^{\frac{\pi}{4}}2^{\sin{x}}g(x)\,dx}$
$\displaystyle{\int_{\frac{\pi}{2}}^{\frac{3\pi}{4}}f(x)\,dx=\int_0^{\frac{\pi}{4}}2^{\sin{(x+\frac{\pi}{4})}}f(x)\,dx}$
$\displaystyle{\int_{\frac{3\pi}{4}}^{\pi}f(x)\,dx=\int_0^{\frac{\pi}{4}}2^{\sin{x}+\cos{x}}g(x)\,dx}$
$\displaystyle{\int_{\pi}^{\frac{5\pi}{4}}f(x)\,dx=\int_0^{\frac{\pi}{4}}2^{\sin{(x+\frac{\pi}{4})}+\cos{(x+\frac{\pi}{4})}}f(x) \,dx}$
$\displaystyle{\int_{\frac{5\pi}{4}}^{\frac{3\pi}{2}}f(x)\,dx=\int_0^{\frac{\pi}{4}}2^{\cos{x}}g(x)\,dx}$
$\displaystyle{\int_{\frac{3\pi}{2}}^{\frac{7\pi}{4}}f(x)\,dx=\int_0^{\frac{\pi}{4}}2^{\cos{(x+\frac{\pi}{4})}}f(x)\,dx}$
$\displaystyle{\int_{\frac{7\pi}{4}}^{2\pi}f(x)\,dx=\int_0^{\frac{\pi}{4}}g(x)\,dx}$
บวกทุกเทอมเข้าด้วยกันจะได้
$\displaystyle{\int_{0}^{2\pi}f(x)\,dx=\int_0^{\frac{\pi}{4}}2\,dx}$
$~~~~~~~~~~~~~~~=\dfrac{\pi}{2}$
แต่สังเกตว่า $f(x)=f(2\pi-x)$ ดังนั้นถ้าแทนค่าตัวแปร $u=2\pi-x$ จะได้ว่า
$\displaystyle{\int_{\pi}^{2\pi}f(x)\,dx=\int_0^{\pi}f(x)\,dx}$
ดังนั้น $\displaystyle{\int_0^{\pi}f(x)\,dx=\dfrac{\pi}{4}}$
สุดท้ายจะได้ว่า
$\displaystyle{\int_0^{7\pi}f(x)\,dx=\dfrac{\pi}{2}+\dfrac{\pi}{2}+\dfrac{\pi}{2}+\dfrac{\pi}{4}}$
$~~~~~~~~~~~~~~~~=\dfrac{7\pi}{4}$
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I'm sorry for my silly solution. I've just found a better approach.
It makes sense to split the integral into 4 parts because the function $f$ has only $4$ components.
$\displaystyle{\int_0^{\frac{\pi}{2}}f(x)\,dx=\int_0^{\frac{\pi}{2}}f(x)\,dx}$
$\displaystyle{\int_{\frac{\pi}{2}}^{\pi}f(x)\,dx=\int_0^{\frac{\pi}{2}}2^{\sin{(x+\frac{\pi}{4})}}f(x)\,dx}$
$\displaystyle{\int_{\pi}^{\frac{3\pi}{2}}f(x)\,dx=\int_0^{\frac{\pi}{2}}2^{\sin{(x+\frac{\pi}{4})}+\cos{(x+\frac{\pi}{4})}}f(x) \,dx}$
$\displaystyle{\int_{\frac{3\pi}{2}}^{2\pi}f(x)\,dx=\int_0^{\frac{\pi}{2}}2^{\sin{(x+\frac{\pi}{4})}}f(x)\,dx}$
บวกทุกเทอมเข้าด้วยกันจะได้
$\displaystyle{\int_{0}^{2\pi}f(x)\,dx=\int_0^{\frac{\pi}{2}}1\,dx}$
$~~~~~~~~~~~~~~~=\dfrac{\pi}{2}$