삼각함수 적분 표
Table of Integrals with Trigonometric Functions
공식
$$ \begin{equation} \int_{0}^{\pi / 2} \sin \theta \cos \theta d \theta = \dfrac{1}{2} \end{equation} $$
$$ \begin{equation} \begin{aligned} \int \cos^{2}\theta d\theta &= \dfrac{1}{2}\theta + \dfrac{1}{4}\sin 2\theta + C \\ \int_{0}^{2\pi} \cos^{2}\theta d\theta &= \pi \\ \int_{0}^{\pi} \cos^{2}\theta d\theta &= \dfrac{\pi}{2} \end{aligned} \end{equation} $$
$$ \begin{equation} \begin{aligned} \int \sin^{2}\theta d\theta &= \dfrac{1}{2}\theta - \dfrac{1}{4}\sin 2\theta + C \\ \int_{0}^{2\pi} \sin^{2}\theta d\theta &= \pi \\ \int_{0}^{\pi} \sin^{2}\theta d\theta &= \dfrac{\pi}{2} \end{aligned} \end{equation} $$
$a \in \mathbb{R}$에 대해서,
$$ \begin{equation} \begin{aligned} \int_{-\infty}^{\infty} \dfrac{\sin (ax)}{x}dx &= \pi \\ \int_{0}^{\infty} \dfrac{\sin (ax)}{x}dx &= \dfrac{\pi}{2} \\ \end{aligned} \end{equation} $$
증명
(1)
$\cos \theta \equiv x$으로 치환하면 적분 요소들은 다음과 같이 바뀐다.
$$ -\sin \theta d\theta = dx \quad \text{and} \quad \int_{\theta=0}^{\pi/2} = \int_{x = 1}^{0} $$
따라서
$$ \int_{0}^{\pi / 2} \sin \theta \cos \theta d \theta = \int_{1}^{0} - x dx = \int_{0}^{1}x dx = \left[ \dfrac{1}{2}x^{2} \right]_{0}^{1} = \dfrac{1}{2} $$
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(2)
삼각함수의 반각공식에 의해, $\cos^{2}\theta = \dfrac{1}{2} + \dfrac{1}{2}\cos 2\theta$이므로,
$$ \begin{align*} \int \cos^{2}\theta d\theta &= \int \left( \dfrac{1}{2} + \dfrac{1}{2}\cos 2\theta \right) d\theta \\ &= \dfrac{1}{2}\theta + \dfrac{1}{4}\sin 2\theta + C \end{align*} $$
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(3)
삼각함수의 반각공식에 의해, $\sin^{2}\theta = \dfrac{1}{2} - \dfrac{1}{2}\cos 2\theta$이므로,
$$ \begin{align*} \int \sin^{2}\theta d\theta &= \int \left( \dfrac{1}{2} - \dfrac{1}{2}\cos 2\theta \right) d\theta \\ &= \dfrac{1}{2}\theta - \dfrac{1}{4}\sin 2\theta + C \end{align*} $$
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(4)
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