삼각함수의 적분 모음 #2

(※별도의 언급이 없으면 $n$은 자연수, $C$는 적분상수를 의미)

삼각함수의 거듭제곱Ⅰ((2n-1)제곱형)

$$\bbox[#FFFFCC,2pt]{21.} \int \sin^{2n-1} x \,dx = \sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}} \frac{(-1)^{k+1}}{2k+1}\cos^{2k+1}x+C$$

<풀이>

$$\begin{align} \int \sin^{2n-1} x \,dx &= \int (1-\cos^2 x)^{n-1} \sin x \,dx\\[5pt]&= \int (1-t^2)^{n-1} \,dt \quad (-\cos x=t\text{로 치환})\\[5pt]&=\int \left( \sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}}(-t^2)^k \right) \,dt\\[5pt]&=\int \left( \sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}} (-1)^k t^{2k} \right)\,dt\\[8pt]&=\sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}} (-1)^k \frac{t^{2k+1}}{2k+1} +C \\[5pt]&=\sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}}\frac{(-1)^{k+1}}{2k+1}\cos^{2k+1}x +C \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{22.} \int \cos^{2n-1} x \,dx =\sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}} \frac{(-1)^k}{2k+1}\sin^{2k+1} x +C $$

<풀이>

$$\begin{align} \int \cos^{2n-1} x \,dx &= \int (1-\sin^2 x)^{n-1} \cos x \,dx \\[5pt]&= \int (1-t^2)^{n-1} \,dt \quad (\sin x=t\text{로 치환})\\[5pt]&=\int \left( \sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}}(-t^2)^k\right)\,dt \\[5pt]&=\int \left( \sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}}(-1)^k t^{2k}\right) \,dt\\[5pt]&=\sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}}(-1)^k \frac{t^{2k+1}}{2k+1}+C\\[5pt]&=\sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}}\frac{(-1)^k}{2k+1}\sin^{2k+1}x +C \quad \blacksquare \end{align}$$

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삼각함수의 거듭제곱Ⅱ(2n제곱형)

$$\bbox[#FFFFCC,2pt]{23.} \int \sec^{2n} x \,dx = \sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}}\frac{\tan^{2k+1}x}{2k+1}+C$$

<풀이>

$$\begin{align} \int \sec^{2n} x \,dx &= \int (\tan^2 x+1)^{n-1} \sec^2 x \,dx \\[5pt]&= \int (t^2 +1)^{n-1} \,dt \quad (\tan x = t \text{로 치환})\\[5pt]&= \int \left( \sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}} t^{2k}\right) \,dt\\[5pt]&= \sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}} \frac{t^{2k+1}}{2k+1}+C\\[5pt]&=\sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}} \frac{\tan^{2k+1}x}{2k+1}+C \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{24.} \int \csc^{2n} x \,dx = -\sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}}\frac{\cot^{2k+1}x}{2k+1}+C$$

<풀이>

$$\begin{align} \int \csc^{2n}x \,dx &= \int (\cot^2 x+1)^{n-1}\csc^2 x \,dx \\[5pt]&=-\int (t^2+1)^{n-1} \,dt \quad (\cot x=t\text{로 치환}) \\[5pt]&= -\int \left( \sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}} t^{2k} \right) \,dt \\[5pt]&= -\sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}} \frac{t^{2k+1}}{2k+1}+C \\[5pt]&= -\sum_{k=0}^{n-1} \sideset{_{n-1}}{_k} {\mathrm{C}} \frac{\cot^{2k+1}x}{2k+1}+C \quad \blacksquare \end{align}$$

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삼각함수의 거듭제곱Ⅲ(점화식)

$$\bbox[#FFFFCC,2pt]{25.} \int \sin^n x \,dx = -\frac{\sin^{n-1}x \cos x}{n}+\frac{n-1}{n}\int \sin^{n-2}x \,dx$$

<풀이>

부분적분법과 $-\cos^2 x=\sin^2 x-1$에 의해

$$\begin{align} \int {\color{#006dd7}\sin^n x} \,dx &= \sin^{n-1}x (-\cos x) -\int (n-1)\sin^{n-2}x (-\cos^2 x)\,dx \\[5pt]&=-\sin^{n-1}x\cos x-(n-1)\int ({\color{#006dd7}\sin^n x}-\sin^{n-2}x)\,dx\end{align}$$

이므로

$$\int {\color{#006dd7}\sin^n x} \,dx = -\frac{\sin^{n-1}x\cos x}{n}+\frac{n-1}{n}\int \sin^{n-2}x\,dx \quad \blacksquare$$

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$$\bbox[#FFFFCC,2pt]{26.} \int \cos^n x \,dx = \frac{\cos^{n-1}x \sin x}{n}+\frac{n-1}{n}\int \cos^{n-2}x\,dx$$

<풀이>

부분적분법과 $-\sin^2 x=\cos^2 x-1$에 의해

$$\begin{align}\int {\color{#006dd7}\cos^n x} \,dx &= \cos^{n-1}x\sin x -\int (n-1)\cos^{n-2}x(-\sin^2 x) \,dx\\[5pt]&= \cos^{n-1}x\sin x -(n-1)\int ({\color{#006dd7}\cos^n x}- \cos^{n-2}x) \,dx \end{align}$$

이므로

$$\begin{align}\int {\color{#006dd7}\cos^n x} \,dx = \frac{\cos^{n-1}x\sin x}{n}+\frac{n-1}{n}\int \cos^{n-2}x \,dx \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{27.} \int \tan^n x \,dx = \frac{\tan^{n-1}x}{n-1} -\int \tan^{n-2}x \,dx \quad (n \neq 1)$$

<풀이>

$$\begin{align}\int \tan^n x \,dx &=\int \tan^{n-2}x(\sec^2 x-1) \,dx \\[5pt]&=\int t^{n-2} \,dt -\int \tan^{n-2}x \,dx\quad (\tan x=t\text{로 치환})\\[5pt]&=\frac{t^{n-1}}{n-1}-\int \tan^{n-2}x \,dx\\[5pt]&= \frac{\tan^{n-1}x}{n-1}-\int \tan^{n-2}x \,dx \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{28.} \int \sec^n x \,dx = \frac{\sec^{n-2}x\tan x}{n-1}+\frac{n-2}{n-1}\int \sec^{n-2}x \,dx \quad (n \neq 1)$$

<풀이>

부분적분법과 $\tan^2 x=\sec^2 x -1$에 의해

$$\begin{align} \int {\color{#006dd7}\sec^n x} \,dx &= \sec^{n-2}x\tan x -\int (n-2)\sec^{n-2}x\tan^2 x \,dx \\[5pt]&=\sec^{n-2}x\tan x -(n-2)\int ({\color{#006dd7}\sec^n x}-\sec^{n-2}x)\,dx \end{align}$$

이므로 

$$\int {\color{#006dd7}\sec^n x} \,dx = \frac{\sec^{n-2}x\tan x}{n-1} +\frac{n-2}{n-1}\int \sec^{n-2}x \,dx \quad \blacksquare$$

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$$\bbox[#FFFFCC,2pt]{29.} \int \cot^n x \,dx = -\frac{\cot^{n-1}x}{n-1} -\int \cot^{n-2}x \,dx \quad (n \neq 1)$$

<풀이>

$$\begin{align} \int \cot^n x \,dx &= \int \cot^{n-2}x(\csc^2 x -1) \,dx \\[5pt]&= -\int t^{n-2}\,dt -\int \cot^{n-2}x \,dx \quad (\cot x =t\text{로 치환}) \\[5pt]&=-\frac{t^{n-1}}{n-1} -\int \cot^{n-2}x \,dx \\[5pt]&=-\frac{\cot^{n-1}x}{n-1} -\int \cot^{n-2}x \,dx \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{30.} \int \csc^n x \,dx = -\frac{\csc^{n-2}x \cot x}{n-1} +\frac{n-2}{n-1} \int \csc^{n-2}x \,dx \quad (n \neq 1)$$

<풀이>

부분적분법과 $\cot^2 x = \csc^2 x -1$에 의해

$$\begin{align} \int {\color{#006dd7}\csc^n x} \,dx &= \csc^{n-2}x(-\cot x)-\int (n-2) \csc^{n-2}x\cot^2 x \,dx \\[5pt]&= -\csc^{n-2}x \cot x -(n-2) \int ({\color{#006dd7}\csc^n x} -\csc^{n-2} x) \,dx \end{align}$$

이므로

$$\int {\color{#006dd7}\csc^n x} \,dx = -\frac{\csc^{n-2}x \cot x}{n-1} +\frac{n-2}{n-1}\int \csc^{n-2}x \,dx \quad \blacksquare$$