삼각함수의 적분

Since the integrand is the product of polynomial and trigonometric functions, we suspect it might yield to integration by parts. There are a number of ways this can be accomplished.

• Solution 1: Before we choose parts, let's use the identity $\sin(2x) = 2\sin x \cos x$.

$$\begin{aligned} \int x\sin x \cos x \mathrm{d}x&=\frac{1}{2}\int x \sin(2x)\mathrm{d}x \\ \text{Now let $u= x$, $\mathrm{d}v=\sin(2x)\mathrm{d}x$; $\mathrm{d}u=\mathrm{d}x$, and $v=-\frac{1}{2}\cos (2x) $. Using integration by parts:} \\ &=\frac{1}{2}\left[-\frac{x}{2}\cos (2x) +\frac{1}{2} \int \cos (2x) \mathrm{d}x\right]\\ &=-\frac{x}{4}\cos (2x) +\frac{1}{8}\sin (2x) +C\\ &=-\frac{x}{4}(1-2\sin^2x) +\frac{1}{4}\sin x\cos x +C\\ &=-\frac{x}{4} + \frac{x}{2}\sin^2x+\frac{1}{4}\sin x \cos x +C \end{aligned}$$

• Solution 2: If we let $u=x$, then $\mathrm{d}u=\mathrm{d}x$, and this seems desirable for integration by parts. If $u=x$, then $\mathrm{d}v = \sin x \cos x \mathrm{d}x$. To find $v$ we can use the substitution $u=\sin x$, $\mathrm{d}u=\cos x \mathrm{d}x$.

$$\begin{aligned} v=\int \sin x \cos x \mathrm{d}x&=\int u \mathrm{d}u = \frac{1}{2}u^2+C = \frac{1}{2}\sin^2 x +C \\ \text{So, we take $v = \frac{1}{2}\sin^2 x$. Now we can apply integration by parts to our original integral.} \\ \int x\sin x \cos x ~\mathrm{d}x&=\frac{x}{2}\sin^2 x - \int \frac{1}{2}\sin^2 x \mathrm{d}x \\ \text{Apply the identity $\sin^2x = \dfrac{1-\cos(2x)}{2}$.} \\ &=\frac{x}{2}\sin^2 x - \frac{1}{4}\int 1-\cos(2 x) \mathrm{d}x\\ &=\frac{x}{2}\sin^2 x - \frac{x}{4} +\frac{1}{8}\sin(2 x)+C\\ &=\frac{x}{2}\sin^2 x - \frac{x}{4} +\frac{1}{4}\sin x \cos x+C \end{aligned}$$

• Solution 3: Let $u=x\sin x$ and $\mathrm{d}v=\cos x \mathrm{d}x$; then $\mathrm{d}u = (x\cos x + \sin x)\mathrm{d}x$ and $v = \sin x$.

$$\begin{aligned} \int x\sin x \cos x \mathrm{d}x&=x\sin^2 x - \int \sin x(x\cos x + \sin x)\mathrm{d}x\\ &=x\sin^2 x - \int x \sin x \cos x\mathrm{d}x - \int \sin^2 x\mathrm{d}x \\ \text{Apply the identity $\sin^2 x = \dfrac{1-\cos(2x)}{2}$ to the second integral.} \\ &=x\sin^2 x - \int x \sin x \cos x\mathrm{d}x - \int \dfrac{1-\cos(2x)}{2}\mathrm{d}x \\&=x\sin^2 x - \int x \sin x \cos x\mathrm{d}x - \frac{x}{2} +\frac{1}{4}\sin(2x)+C \\ \text{So, we have the equation} \\ \int x\sin x \cos x \mathrm{d}x&=x\sin^2 x - \int x \sin x \cos x\mathrm{d}x - \frac{x}{2} + \frac{1}{4}\sin(2x)+C\\ 2\int x\sin x \cos x \mathrm{d}x&=x\sin^2 x - \frac{x}{2} + \frac{1}{4}\sin(2x)+C\\ \int x\sin x \cos x \mathrm{d}x&=\frac{x}{2}\sin^2 x - \frac{x}{4} + \frac{1}{8}\sin(2x)+\frac{C}{2}\\ &=\frac{x}{2}\sin^2 x - \frac{x}{4} + \frac{1}{4}\sin x\cos x+\frac{C}{2} \end{aligned}$$

Since $C$ is an arbitrary constant that can take any number in $(-\infty,\infty)$, also $\frac{C}{2}$ is an arbitrary constant that can take any number in $(-\infty,\infty)$, so we're free to rename $\frac{C}{2}$ to $C$.