삼각함수의 적분

Recall $\tan x = \dfrac{\sin x}{\cos x}$.

$$\begin{aligned} \int \tan^5 x \cos^2 x ~\mathrm{d}x&=\int \frac{\sin^5 x}{\cos^5 x}\cos^2 x ~\mathrm{d}x =\int \frac{\sin^5 x}{\cos^3 x}~\mathrm{d}x \\ \text{Substitute $u=\cos x$, so $\mathrm{d}u=-\sin x~\mathrm{d}x$ and $\sin^2 x = 1-\cos^2 x = 1-u^2$.} \\ &=\int \frac{\sin^4 x}{\cos^3 x}~\sin x~\mathrm{d}x =-\int \frac{(1-u^2)^2}{u^3}~\mathrm{d}u\\ &=-\int \frac{1-2u^2+u^4}{u^3}~\mathrm{d}u= \int \left(-\frac{1}{u^3}+\frac{2}{u}-u\right)\mathrm{d}u\\ &=\frac{1}{2u^2}+2\log|u|-\frac{1}{2}u^2+C\\ &=\frac{1}{2\cos^2 x}+2\log|\cos x|-\frac{1}{2}\cos^2 x +C \end{aligned}$$

We can use the definition of secant to make this integral look more familiar. $$\int \frac{1}{\cos^2 \theta}~\mathrm{d}\theta = \int \sec^2\theta~\mathrm{d}\theta = \tan \theta +C$$

We re-write $\cot x = \dfrac{\cos x}{\sin x}$, and use the substitution $u=\sin x$, $\mathrm{d}u=\cos x~\mathrm{d}x$.

$$\begin{aligned} \int \cot x~\mathrm{d}x&= \int \frac{\cos x}{\sin x}~\mathrm{d}x = \int \frac{1}{u}~\mathrm{d}u\\ &=\log|u|+C = \log|\sin x|+C \end{aligned}$$

• Solution 1: We begin with the obvious substitution, $w=e^x$, $\mathrm{d}w=e^x \mathrm{d}w$.

$$\begin{aligned} \int e^x\sin(e^x)\cos(e^x) ~\mathrm{d}x&= \int \sin w \cos w ~\mathrm{d}w \\ \text{Now we see another substitution, $u=\sin w$, $\mathrm{d}u=\cos w~\mathrm{d}w$.} \\ &=\int u~\mathrm{d}u=\frac{1}{2}u^2+C=\frac{1}{2}\sin^2 w +C\\ &=\frac{1}{2}\sin^2(e^x)+C \end{aligned}$$

• Solution 2: Notice that $\frac{\mathrm{d}}{\mathrm{d}x}\{\sin(e^x)\} = e^x \cos(e^x)$. This suggests to us the substitution $u=\sin(e^x)$, $\mathrm{d}u = e^x \cos(e^x)~\mathrm{d}x$.

$$\begin{aligned} \int e^x\sin(e^x)\cos(e^x) ~\mathrm{d}x&= \int u~\mathrm{d}u =\frac{1}{2}u^2+C = \frac{1}{2}\sin^2(e^x)+C \end{aligned}$$

Since we have an “inside function," we start with the substitution $s=\cos x$, so $-\mathrm{d}s=\sin x ~\mathrm{d}x$ and $\sin^2 x = 1-\cos^2 x = 1-s^2$.

$$\begin{aligned} \int \sin(\cos x)\sin^3 x ~\mathrm{d}x&=\int \sin(\cos x) \cdot \sin^2 x \cdot \sin x \mathrm{d}x\\ &=-\int \sin(s)\cdot (1-s^2) ~\mathrm{d}s \\ \text{We use integration by parts with $u=(1-s^2)$, $\mathrm{d}v=\sin s ~\mathrm{d}s$; $\mathrm{d}u=-2s~\mathrm{d}s$, and $v = -\cos s$.} \\ &=-\left[-(1-s^2)\cos s - \int 2s\cos s~ \mathrm{d}s\right] \\&= (1-s^2)\cos s + \int 2s\cos s~ \mathrm{d}s \\ \text{We integrate by parts again, with $u=2s$, $\mathrm{d}v=\cos s ~\mathrm{d}s$; $\mathrm{d}u=2~\mathrm{d}s$, and $v=\sin s$.} \\ &=(1-s^2)\cos s + 2s\sin s - \int 2\sin s~\mathrm{d}s \\&=(1-s^2)\cos s + 2s\sin s +2\cos s +C \\&=\sin^2 x\cdot\cos (\cos x) + 2\cos x\cdot\sin (\cos x) +2\cos (\cos x) +C \\&=(\sin^2x+2)\cos (\cos x) + 2\cos x\cdot\sin (\cos x) +C \end{aligned}$$

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 $C$.