MATH FACTORY

$x = 2 \sin \theta \ \left( - \dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2} \right)$라 하면 $x=0$일 때 $\theta = 0$, $x=2$일 때 $\theta = \dfrac{\pi}{2}$이고, 양변을 $x$로 미분하면

$$\begin{gather*} 1 = 2 \cos \theta \dfrac{d \theta}{dx} \ \ \ \therefore \ \ dx = 2 \cos \theta d \theta \end{gather*}$$

따라서

$$\begin{align*} \int_{0}^{2} \sqrt{4 - x^2} dx &= \int_{0}^{\dfrac{\pi}{2}} \sqrt{4 - 4 \sin^2 \theta} \cdot 2 \cos \theta d\theta \\ &= \int_{0}^{\dfrac{\pi}{2}} \sqrt{4 \cos^2 \theta} \cdot 2 \cos \theta d\theta \\ &= \int_{0}^{\dfrac{\pi}{2}} 4 \cos^2 \theta d\theta \\ &= 2 \int_{0}^{\dfrac{\pi}{2}} (1 + \cos 2\theta ) d\theta \\ &= 2 \left[ \theta + \dfrac{1}{2} \sin 2 \theta \right]_{0}^{\dfrac{\pi}{2}} = \pi \end{align*}$$

$x = \sin \theta \ \left( - \dfrac{\pi}{2} \leq \theta \leq \dfrac{\pi}{2} \right)$라 하면 $x=0$일 때 $\theta = 0$, $x=\dfrac{\sqrt{3}}{2}$일 때 $\theta = \dfrac{\pi}{3}$이고, 양변을 $x$로 미분하면

$$\begin{gather*} 1 = \cos \theta \dfrac{d \theta}{dx} \ \ \ \therefore \ \ dx = \cos \theta d \theta \end{gather*}$$

따라서

$$\begin{align*} \int_{0}^{\frac{\sqrt{3}}{2}} \dfrac{1}{\sqrt{1 - x^2}} dx &= \int_{0}^{\frac{\pi}{3}} \dfrac{1}{\sqrt{1 - \sin^2 \theta}} \cdot \cos \theta d\theta \\ &= \int_{0}^{\frac{\pi}{3}} \dfrac{1}{\cos \theta} \cdot \cos \theta d\theta \\ &= \int_{0}^{\frac{\pi}{3}} d\theta = \Big[ \theta \Big]_{0}^{\dfrac{\pi}{3}} = \dfrac{\pi}{3} \end{align*}$$

$x = 2 \tan \theta \ \left( - \dfrac{\pi}{2} < \theta < \dfrac{\pi}{2} \right)$라 하면 $x=0$일 때 $\theta = 0$, $x=2$일 때 $\theta = \dfrac{\pi}{4}$이고, 양변을 $x$로 미분하면

$$\begin{gather*} 1 = 2 \sec^2 \theta \dfrac{d \theta}{dx} \ \ \ \therefore \ \ dx = 2 \sec^2 \theta d\theta \end{gather*}$$

따라서

$$\begin{align*} \int_{0}^{2} \dfrac{1}{x^2 + 4} dx &= \int_{0}^{\frac{\pi}{4}} \dfrac{2\sec^2 \theta}{4 \tan^2 \theta + 4} dx \\ &= \int_{0}^{\frac{\pi}{4}} \dfrac{2\sec^2 \theta}{4 \sec^2 \theta} dx \\ &= \int_{0}^{\frac{\pi}{4}} \dfrac{1}{2} d\theta \\ &= \left[ \dfrac{1}{2} \theta \right]_{0}^{\frac{\pi}{4}} = \dfrac{\pi}{8} \end{align*}$$