미적분의 기본정리

• Solution 1: This might not obviously look like the derivative of anything familiar, but it does look like half of a familiar trig identity: $2\sin x \cos x = \sin(2x)$.

$$\begin{aligned} \int 3 \sin x \cos x \ \mathrm{d}x&=\int \frac{3}{2}\cdot 2\sin x \cos x \ \mathrm{d}x\\ &=\int \frac{3}{2} \sin(2x)\ \mathrm{d}x \\ \text{So, we might guess that the antiderivative is something like $-\cos(2x)$. We only need to figure out the constants.} \\ \frac{\mathrm{d}}{\mathrm{d}x}\{-\cos(2x)\}&=2\sin(2x)\\ \text{So,}\qquad \frac{\mathrm{d}}{\mathrm{d}x}\left\{-\frac{3}{4}\cos(2x)\right\}&=\frac{3}{2}\sin(2x)\\ \text{Therefore,}\qquad \int 3\sin x \cos x\ \mathrm{d}x&=-\frac{3}{4}\cos(2x)+C \end{aligned}$$

• Solution 2: You might notice that the integrand looks like it came from the chain rule, since $\cos x$ is the derivative of $\sin x$. Using this observation, we can work out the antideriative:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\left\{\sin^2 x\right\}&=2\sin x \cos x\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{3}{2}\sin^2 x\right\}&=3\sin x \cos x\\ \text{So,}\qquad \int 3\sin x \cos x \ \mathrm{d}x&=\frac{3}{2}\sin^2 x+C \end{aligned}$$

These two answers look different. Using the identity $\cos(2x)=1 - 2 \sin^2(x)$, we reconcile them:

$$\begin{aligned} -\frac{3}{4}\cos(2x)+C&= -\frac{3}{4}\left(1-2\sin^2x\right)+C\\ &=\frac{3}{2}\sin^2 x + \left(C-\frac{3}{4}\right) \end{aligned}$$

The $\frac{3}{4}$ here is not significant. Remember that $C$ is used to designate a constant that can take any value between $-\infty$ and $+\infty$. So $C-\frac{3}{4}$ is also just a constant that can take any value between $-\infty$ and $+\infty$. As the two answers we found differ by a constant, they are equivalent.