부정적분의 뜻과 계산

We differentiate $f(x)$ and $g(x)$ using the chain rule. Recall $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\arcsin(x)\right\}=\dfrac{1}{\sqrt{1-x^2}}$.

$$\begin{aligned} f'(x)&=\frac{2}{\sqrt{1- \sqrt{x}^2}}\cdot\frac{\mathrm{d}}{\mathrm{d}x} \sqrt{x}\\ &=\frac{2}{\sqrt{1-x}}\cdot\frac{1}{2\sqrt{x}}\\ &=\frac{1}{\sqrt{x-x^2}}\\ g'(x)&=\frac{1}{\sqrt{1- (2x-1)^2}}\cdot\frac{\mathrm{d}}{\mathrm{d}x}\{ 2x-1\}\\ &=\frac{2}{\sqrt{1-4x^2+4x-1}}\\ &=\frac{1}{\sqrt{x-x^2}}\\ \end{aligned}$$

So, the derivative of $f(x)-g(x)$ is $0$, which implies that $f(x)$ and $g(x)$ differ by a constant–perhaps a surprising result!

Remark: we don't need calculus to show that $f(x)$ and $g(x)$ only differ by a constant. Define $\theta=\sin^{-1}\sqrt{x}$, so that $\sin\theta=\sqrt{x}$ and $f(x) = 2\theta$ , and then $$\sin \left[f(x)-\frac{\pi}{2}\right]=-\cos f(x) =-\cos 2\theta=-[1-2\sin^2\theta] =2\sin^2\theta -1=2x-1$$ Our goal is to take the arcsine of the first and last expressions and conclude $f(x)-\dfrac{\pi}{2}=\arcsin(2x-1)=g(x)$. However, before we can say $\arcsin\left(\sin\left[f(x)-\dfrac{\pi}{2}\right]\right)=f(x)-\dfrac{\pi}{2}$, we have to check that $-\dfrac{\pi}{2} \leq f(x)-\dfrac{\pi}{2}\leq\dfrac{\pi}{2}$. That is, we need to show that $0 \leq f(x) \leq \pi$.

As $0\le\theta\le\frac{\pi}{2}$ (since $\sin^{-1}$ always takes values between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ and since $\sqrt{x}\ge 0$), we have that $0\le f(x)=2\theta\le \pi$. So, $$f(x)-\dfrac{\pi}{2}=\arcsin\left(\sin\left[f(x)-\dfrac{\pi}{2}\right]\right)=\arcsin(2x+1)=g(x).$$