삼각함수의 적분 모음 #1

 (※본 포스트에서 $C$는 적분상수를 의미)

기본 공식(미분 공식의 역방향)

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

 

$$\bbox[#FFFFCC,2pt]{2.} \int \cos x \,dx= \sin x + C$$

 

$$\bbox[#FFFFCC,2pt]{3.} \int \sec^2 x \,dx = \tan x + C$$

 

$$\bbox[#FFFFCC,2pt]{4.}  \int \sec x \tan x \,dx = \sec x +C$$

 

$$\bbox[#FFFFCC,2pt]{5.}  \int \csc^2 x \,dx = -\cot x +C$$

 

$$\bbox[#FFFFCC,2pt]{6.}  \int \csc x \cot x \,dx = -\csc x +C$$

 

 

삼각함수의 기본형(중복 제외)

$$\bbox[#FFFFCC,2pt]{7.} \int \tan x \,dx= \ln \left| \sec x \right| +C$$

<풀이>

$$\begin{align}\int \tan x \,dx &=\int \frac{\sin x}{\cos x}\,dx\\[5pt]&=\int \frac{-1}{t}\,dt\quad(\cos x=t\text{로 치환})\\[11pt]&=-\ln \left| t \right|+C\\[17pt]&=\ln \left| \sec x \right| +C \quad\blacksquare\end{align}$$

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$$\bbox[#FFFFCC,2pt]{8.} \int \sec x \,dx = \ln \left| \sec x +\tan x \right| +C$$

<풀이>

$$\begin{align} \int \sec x \,dx &=\int \frac{\cos x}{1-\sin^2 x}\,dx\\[5pt]&=\int \frac{1}{1-t^2}\,dt \quad(\sin x=t \text{로 치환})\\[5pt]&=\frac{1}{2}\int \left( \frac{1}{t+1}-\frac{1}{t-1}\right)\,dt\\[8pt]&=\frac{1}{2}\left( \ln\left| t+1\right|-\ln\left| t-1\right|\right)+C\\[11pt]&=\frac{1}{2}\ln\left| \frac{\sin x+1}{\sin x-1}\right|+C\\[11pt]&=\frac{1}{2}\ln \frac{(\sin x+1)^2}{\cos^2 x}+C\\[8pt]&=\ln\left| \frac{\sin x+1}{\cos x}\right|+C\\[14pt]&=\ln \left| \sec x+\tan x\right|+C \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{9.} \int \cot x \,dx = -\ln\left| \csc x \right| +C$$

<풀이>

$$\begin{align}\int \cot x \,dx &= \int \frac{\cos x}{\sin x} \,dx\\[5pt]&=\int \frac{1}{t}\,dt\quad(\sin x=t\text{로 치환})\\[11pt]&=\ln\left|\sin x\right|+C\\[17pt]&=-\ln\left|\csc x\right| +C \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{10.} \int \csc x \,dx = -\ln \left| \csc x+\cot x\right|+C$$

<풀이>

$$\begin{align}\int \csc x \,dx &= \int \frac{\sin x}{1-\cos^2 x}\,dx\\[5pt]&=\int \frac{-1}{1-t^2}\,dt\quad(\cos x=t\text{로 치환})\\[5pt]&=\frac{1}{2}\int \left( \frac{1}{t-1}-\frac{1}{t+1}\right)\,dt\\[8pt]&=\frac{1}{2}\left( \ln\left| t-1\right|-\ln\left| t+1\right|\right)+C\\[11pt]&=-\frac{1}{2}\ln\left| \frac{\cos x+1}{\cos x-1}\right|+C \\[11pt]&=-\frac{1}{2}\ln\frac{(\cos x+1)^2}{\sin^2 x}+C\\[11pt]&=-\ln\left|\frac{\cos x+1}{\sin x} \right|+C\\[17pt]&=-\ln\left|\csc x+\cot x \right|+C \quad \blacksquare \end{align}$$

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삼각함수의 제곱형(중복 제외)

$$\bbox[#FFFFCC,2pt]{11.} \int \sin^2 x \,dx = \frac{x}{2}-\frac{\sin 2x}{4}+C$$

<풀이>

$$\begin{align}\int \sin^2 x \,dx &= \int \frac{1-\cos 2x}{2}\,dx\\[5pt]&=\frac{x}{2}-\frac{\sin 2x}{4}+C \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{12.} \int \cos^2 x \,dx = \frac{x}{2}+\frac{\sin 2x}{4}+C$$

<풀이>

$$\begin{align} \int \cos^2 x \,dx &= \int \frac{1+\cos 2x}{2} \,dx\\[5pt]&=\frac{x}{2}+\frac{\sin 2x}{4}+C \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{13.} \int \tan^2 x \,dx = \tan x-x+C$$

<풀이>

$$\begin{align} \int \tan^2 x \,dx &= \int (\sec^2 x -1)\,dx\\[5pt]&=\tan x -x +C \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{14.} \int \cot^2 x\,dx = -\cot x-x+C$$

<풀이>

$$\begin{align} \int \cot^2 x \,dx &= \int (\csc^2 x -1) \,dx\\[5pt]&=-\cot x -x +C \quad \blacksquare \end{align}$$

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삼각함수의 세제곱형

$$\bbox[#FFFFCC,2pt]{15.}\int \sin^3 x \,dx=-\cos x+\frac{\cos^3 x}{3} +C$$

<풀이>

$$\begin{align}\int \sin^3 x \,dx &=\int \sin x(1-\cos^2 x) \,dx\\[5pt]&=\int -(1-t^2) \,dt \quad (\cos x=t\text{로 치환})\\[8pt]&=-t+\frac{t^3}{3}+C\\[8pt]&=-\cos x+\frac{\cos^3 x}{3}+C \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{16.} \int \cos^3 x \,dx =\sin x-\frac{\sin^3 x}{3}+C$$

<풀이>

$$\begin{align} \int \cos^3 x \,dx &= \int \cos x(1-\sin^2 x) \,dx \\[5pt]&=\int (1-t^2) \,dt \quad (\sin x=t\text{로 치환})\\[8pt]&=t-\frac{t^3}{3}+C\\[8pt]&=\sin x-\frac{\sin^3 x}{3}+C \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{17.} \int \tan^3 x \,dx = \frac{\tan^2 x}{2}-\ln \left| \sec x \right| +C$$

<풀이>

$$\begin{align} \int \tan^3 x \,dx &=\int \tan x(\sec^2 x-1) \,dx\\[5pt]&=\int t \,dt-\int \tan x \,dx \quad (\tan x=t\text{로 치환})\\[8pt]&=\frac{\tan^2 x}{2}-\ln\left| \sec x \right|+C\quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{18.} \int \sec^3 x \,dx = \frac{\sec x \tan x}{2}+\frac{1}{2}\ln\left| \sec x+\tan x\right|+C$$

<풀이>

부분적분법에 의해

$$\begin{align} \int \sec^3 x \,dx &= \sec x \tan x -\int \sec x \tan^2 x \,dx\\[5pt]&=\sec x\tan x-\int (\sec^3 x-\sec x)\,dx \end{align}$$

이므로

$$\begin{align} \int \sec^3 x \,dx &=\frac{\sec x \tan x}{2}+\frac{1}{2}\int \sec x \,dx\\[5pt]&=\frac{\sec x \tan x}{2}+\frac{1}{2}\ln \left| \sec x + \tan x \right| \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{19.} \int \cot^3 x \,dx = -\frac{\cot^2 x}{2}+\ln \left| \csc x \right| +C$$

<풀이>

$$\begin{align}\int \cot^3 x \,dx &= \int \cot x(\csc^2 x -1) \,dx\\[5pt]&=\int -t \,dt -\int \cot x \,dx\quad (\cot x=t\text{로 치환})\\[5pt]&=-\frac{\cot^2 x}{2}+\ln \left| \csc x \right| +C \quad \blacksquare \end{align}$$

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$$\bbox[#FFFFCC,2pt]{20.} \int \csc^3 \,dx = -\frac{\csc x \cot x}{2}-\frac{1}{2}\ln\left| \csc x+ \cot x \right|+C$$

<풀이>

부분적분법에 의해 

$$\begin{align} \int \csc^3 x \,dx &= -\csc x \cot x -\int \csc x \cot^2 x \,dx\\[5pt]&=-\csc x \cot x -\int (\csc^3 x -\csc x)\,dx \end{align}$$

이므로

$$\begin{align} \int \csc^3 x \,dx &= -\frac{\csc x \cot x}{2}+\frac{1}{2}\int \csc x \,dx\\[5pt]&=-\frac{\csc x \cot x}{2}-\frac{1}{2}\ln\left| \csc x + \cot x \right| +C \quad \blacksquare \end{align}$$