미적분의 기본정리

The integrand is similar to $\dfrac{1}{\sqrt{1-x^2}}$. In order to formulate a guess for the antiderivative, let's factor out $\sqrt{2}$ from the denominator:

$$\begin{aligned} \displaystyle\int \dfrac{1}{\sqrt{2-x^2}}\mathrm{d}x&= \displaystyle\int \dfrac{1}{\sqrt{2\left(1-\frac{x^2}{2}\right)}}\mathrm{d}x\\ &=\displaystyle\int \dfrac{1}{ \sqrt{2}\sqrt{1-\frac{x^2}{2}}}\mathrm{d}x\\ &= \int\frac{1}{\sqrt{2}}\cdot \dfrac{1}{ \sqrt{1-\left(\dfrac{x}{\sqrt2}\right)^2}}\mathrm{d}x\\ \\ \text{At this point, we might guess that our antiderivative is something like $F(x) = \arcsin\left(\dfrac{x}{\sqrt{2}}\right)$. To explore this possibility, we can differentiate, and see what we get.} \\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\arcsin\left(\dfrac{x}{\sqrt{2}}\right)\right\}&=\frac{1}{\sqrt{2}}\cdot\dfrac{1}{\sqrt{1-\left(\dfrac{x}{\sqrt{2}}\right)^2}} \\ \text{This is exactly what we want! So,} \\ \int \dfrac{1}{\sqrt{2-x^2}}\mathrm{d}x&=\arcsin\left(\dfrac{x}{\sqrt2}\right)+C \end{aligned}$$