眞趣盎然集

Integral of x/(36-x^2)^(1/2) from 0 to 3 求 $\int_0^3 \frac{x}{\sqrt{36-x^2}}dx$ 解： $$I=\int_0^3 \frac{x}{\sqrt{36-x^2}}\;dx$$ $$\;\;\;\;\;\;\;=\int_0^3 \frac{x}{\sqrt{36(1- \frac{x^2}{36})}}\;dx$$ $$\;\;\;\;\;\;=\int_0^3 \frac{x}{6 \sqrt{1-(\frac{x}{6})^2}}\;dx$$ 利用 變數變換 (Integration by Substitution) ， 令  $$\frac{x}{6} = \sin\theta$$ 可知  $$\theta = \sin^{-1}\frac{x}{6}$$ 且 $$x=6\sin\theta$$ $$dx=6\cos\theta\; d\theta$$ 則 $$\;\;\;\;\;I=\int_{\sin^{-1}0}^{\sin^{-1}\frac{1}{2}} \frac{(6\sin\theta)(6\cos\theta) }{6\sqrt{1-\sin^2\theta}}\;d\theta$$ $$\;\;\;\;=\int_0^{\frac{\pi}{6}}\frac{(6\sin\theta)(6\cos\theta)}{6\cos\theta}\;d\theta\;\;\;$$ $$\;\;=\int_0^{\frac{\pi}{6}}6\sin\theta\; d\theta\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$$ $$=-6\int_0^{\frac{\pi}{6}}d(\cos\theta)\;\;\;\;\;\;\;\;\;\;\;\;\;$$ $$=\left. -6\cos\theta \right\rvert_0^{\frac{\pi}{6}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$$ $$\;\;=6(\cos 0-\cos \frac{\pi}{6})\;\;\;\;\;\;\;\;\;\;\;\;\;\;$$ $$=6\left (1-\frac{\sq...