삼각함수의 적분 모음 #3

 

역삼각함수

$$\bbox[#FFFFCC,2pt]{31.} \int \arcsin x \,dx = x\arcsin x + \sqrt{1-x^2}+C$$

<풀이>

$\arcsin x = t$라고 하면
$x=\sin t$, $dx = \cos t \,dt$이므로

$$\begin{align} \int \arcsin x \,dx &= \int t \cos t \,dt \\[5pt]&= t\sin t -\int \sin t\,dt \\[5pt]&= t\sin t +{\color{#006dd7}\cos t} +C \\[5pt]&= x\arcsin x + {\color{#006dd7}\sqrt{1-x^2}} +C \quad \blacksquare \end{align}$$

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<참고>

$t=\arcsin x$로부터 $-\frac{\pi}{2} \leq t \leq \frac{\pi}{2}$이기에
$\cos t \geq 0$이므로

$$\begin{align} {\color{#006dd7}\cos t} &= \sqrt{1-\sin^2 t}\\[5pt]&={\color{#006dd7}\sqrt{1-x^2}} \end{align}$$

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$$\bbox[#FFFFCC,2pt]{32.} \int \arccos x \,dx = x\arccos x - \sqrt{1-x^2}+C$$

<풀이>

$\arccos x = t$라고 하면
$x=\cos t$, $dx = -\sin t \,dt$이므로

$$\begin{align} \int \arccos x \,dx &= \int t(-\sin t) \,dt\\[5pt]&= t\cos t -\int \cos t \,dt\\[5pt]&= t\cos t - {\color{#006dd7}\sin t} +C \\[5pt]&= x\arccos x - {\color{#006dd7}\sqrt{1-x^2}}+C \quad \blacksquare \end{align}$$

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<참고>

$t= \arccos x$로부터 $0 \leq t \leq \pi$이기에
$\sin t \geq 0$이므로

$$\begin{align} {\color{#006dd7}\sin t} &= \sqrt{1-\cos^2 t}\\[5pt]&= {\color{#006dd7}\sqrt{1-x^2}} \end{align}$$

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$$\bbox[#FFFFCC,2pt]{33.} \int \arctan x \,dx = x\arctan x - \frac{1}{2}\ln(1+x^2) +C$$

<풀이>

$\arctan x =t$라고 하면
$x=\tan t$, $dx = \sec^2 t\,dt$이므로

$$\begin{align} \int \arctan x \,dx &= \int t \sec^2 t\,dt \\[5pt]&= t\tan t - \int \tan t\,dt \\[5pt]&= t\tan t - \ln\left| {\color{#006dd7}\sec t} \right| +C \\[5pt]&= x\arctan x -\ln {\color{#006dd7}\sqrt{1+x^2}} +C \\[5pt]&=x\arctan x -\frac{1}{2}\ln(1+x^2)+C \quad \blacksquare \end{align}$$

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<참고>

$t=\arctan x$로부터 $-\frac{\pi}{2} < t < \frac{\pi}{2}$이기에
$\sec t >0$이므로

$$\begin{align} {\color{#006dd7}\sec t} &= \sqrt{1+\tan^2 t}\\[5pt]&= {\color{#006dd7}\sqrt{1+x^2}} \end{align}$$

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혼합형Ⅰ

홀수제곱을 포함하는 경우

$$\bbox[#FFFFCC,2pt]{34.} \int \sin x \cos x \,dx = -\frac{1}{4} \cos 2x +C$$

<풀이>

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

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<다른 풀이 1>

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

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<다른 풀이 2>

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

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$$\bbox[#FFFFCC,2pt]{35.} \int \sin^n x \cos x \,dx =\frac{\sin^{n+1}x}{n+1}+C $$

<풀이>

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

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$$\bbox[#FFFFCC,2pt]{36.} \int \sin x \cos^m x \,dx =-\frac{\cos^{m+1}x}{m+1}+C $$

<풀이>

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

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$$\bbox[#FFFFCC,2pt]{37.} \int \sin^{2n+1}x \cos^m x \,dx = \sum_{k=0}^n \sideset{_n}{_k} {\mathrm{C}} \frac{(-1)^{k+1} \cos^{2k+m+1}x}{2k+m+1}+C$$

<풀이>

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

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$$\bbox[#FFFFCC,2pt]{38.} \int \sin^n x \cos^{2m+1}x \,dx = \sum_{k=0}^m \sideset{_m}{_k} {\mathrm{C}} \frac{(-1)^k \sin^{2k+n+1}x}{2k+n+1} +C$$

<풀이>

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

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홀수제곱을 포함하지 않는 경우(점화식)

$$\bbox[#FFFFCC,2pt]{39.} \int \sin^n x \cos^m x \,dx = -\frac{\sin^{n-1}x \cos^{m+1}x}{n+m}+\frac{n-1}{n+m}\int \sin^{n-2}x \cos^m x \,dx$$

<풀이: $\sin^n x$의 지수 줄이기>

부분적분법과 $\cos^2 x = 1 - \sin^2 x$에 의해

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

이므로

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

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$$\bbox[#FFFFCC,2pt]{40.} \int \sin^n x \cos^m x \,dx = \frac{\sin^{n+1}x \cos^{m-1}x}{n+m}+\frac{m-1}{n+m}\int \sin^{n}x \cos^{m-2} x \,dx$$

<풀이: $\cos^m x$의 지수 줄이기>

부분적분법과 $\sin^2 x = 1 - \cos^2 x$에 의해

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

이므로

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

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