📂Lemmas

Integrals of Trigonometric Functions Table

Formulas

$$\begin{equation} \int_{0}^{\pi / 2} \sin \theta \cos \theta d \theta = \dfrac{1}{2} \end{equation}$$

$$\begin{equation} \begin{aligned} \int \cos^{2}\theta d\theta &= \dfrac{1}{2}\theta + \dfrac{1}{4}\sin 2\theta + C \\ \int_{0}^{2\pi} \cos^{2}\theta d\theta &= \pi \\ \int_{0}^{\pi} \cos^{2}\theta d\theta &= \dfrac{\pi}{2} \end{aligned} \end{equation}$$

$$\begin{equation} \begin{aligned} \int \sin^{2}\theta d\theta &= \dfrac{1}{2}\theta - \dfrac{1}{4}\sin 2\theta + C \\ \int_{0}^{2\pi} \sin^{2}\theta d\theta &= \pi \\ \int_{0}^{\pi} \sin^{2}\theta d\theta &= \dfrac{\pi}{2} \end{aligned} \end{equation}$$

Regarding $a \in \mathbb{R}$,

$$\begin{equation} \begin{aligned} \int_{-\infty}^{\infty} \dfrac{\sin (ax)}{x}dx &= \pi \\ \int_{0}^{\infty} \dfrac{\sin (ax)}{x}dx &= \dfrac{\pi}{2} \\ \end{aligned} \end{equation}$$

Proof

(1)

By substituting $\cos \theta \equiv x$, the integral elements are changed as follows.

$$-\sin \theta d\theta = dx \quad \text{and} \quad \int_{\theta=0}^{\pi/2} = \int_{x = 1}^{0}$$

Therefore,

$$\int_{0}^{\pi / 2} \sin \theta \cos \theta d \theta = \int_{1}^{0} - x dx = \int_{0}^{1}x dx = \left[ \dfrac{1}{2}x^{2} \right]_{0}^{1} = \dfrac{1}{2}$$

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(2)

By the half-angle formula of trigonometric functions, since $\cos^{2}\theta = \dfrac{1}{2} + \dfrac{1}{2}\cos 2\theta$,

$$\begin{align*} \int \cos^{2}\theta d\theta &= \int \left( \dfrac{1}{2} + \dfrac{1}{2}\cos 2\theta \right) d\theta \\ &= \dfrac{1}{2}\theta + \dfrac{1}{4}\sin 2\theta + C \end{align*}$$

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(3)

By the half-angle formula of trigonometric functions, since $\sin^{2}\theta = \dfrac{1}{2} - \dfrac{1}{2}\cos 2\theta$,

$$\begin{align*} \int \sin^{2}\theta d\theta &= \int \left( \dfrac{1}{2} - \dfrac{1}{2}\cos 2\theta \right) d\theta \\ &= \dfrac{1}{2}\theta - \dfrac{1}{4}\sin 2\theta + C \end{align*}$$

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(4)

Go to proof

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