อ้างอิง:
ข้อความเดิมเขียนโดยคุณ Timestopper_STG
23.จงแสดงว่าลำดับ $n\sin n$ ไม่มีขอบเขต
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Consider $n\in \left[2k\pi+\dfrac{\pi}{6},2k\pi+\dfrac{5\pi}{6}\right],\forall k\in \mathbb{N} $ we'll get $n\sin n>\dfrac{n}{2}$
So when $\lim(k\rightarrow\infty)$ then $n\sin n>\dfrac{n}{2}=+\infty,\exists n\in\mathbb{N}$ which is diverges
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$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x-b\sin x}{a\sin x+b\cos x}dx=\ln\left(\frac{a}{b}\right)$$
BUT
$$\int_{0}^{\frac{\pi}{2}}\frac{a\cos x+b\sin x}{a\sin x+b\cos x}dx=\frac{\pi ab}{a^{2}+b^{2}}+\frac{a^{2}-b^{2}}{a^{2}+b^{2}}\ln\left(\frac{a}{b}\right)$$