Integration of Trigonometric Functions

Formula 1:

$$\int \sin x \, dx = -\cos x+C $$

Proof:

\begin{align*} \because \frac{d}{dx} \, \cos x &= -\sin x \\\\ \therefore \frac{d}{dx} \, (-\cos x) &= \sin x \\\\ \frac{d}{dx} \, (-\cos x + C) &= \sin x \\\\ \therefore \int \sin x \,dx &= -\cos x +C \\\\ \end{align*}

Formula 2:

$$\int \cos x \, dx = \sin x+C $$

Proof:

\begin{align*} \because \frac{d}{dx} \, \sin x &= \cos x \\\\ \frac{d}{dx} \, (\sin x + C) &= \cos x \\\\ \therefore \int \cos x \,dx &= \sin x +C \\\\ \end{align*}

Formula 3:

$$\int \tan x \, dx = \ln \lvert \sec x \lvert+C$$

Proof:

\begin{align*} \int \tan x \, dx &= \int \frac{\sin x }{\cos x }\, dx\\\\ \textup{Let}\,\, u&=\cos x\\ du &= -\sin x dx\\ -du &= \sin x dx\\\\ \int \tan x\, dx &= \int \frac{-du}{u}\\ &=- \int \frac{1}{u}\,du\\ &= -\ln \lvert u \lvert+C\\ &= -\ln \lvert \cos x \lvert+C\\ &= \ln 1 -\ln \lvert \cos x \lvert+C\\ &= \ln \frac{1}{\lvert \cos x \lvert}+C\\ &= \ln \lvert \sec x \lvert+C\\ \end{align*}

Formula 4:

$$\int \cot x \, dx = \ln \lvert \sin x \lvert+C$$

Proof:

\begin{align*} \int \cot x \, dx &= \int \frac{\cos x }{\sin x }\, dx\\\\ \textup{Let}\,\, u&=\sin x\\ du &= \cos x dx\\\\ \int \cot x\, dx &= \int \frac{du}{u}\\ &= \ln \lvert u \lvert+C\\ &= \ln \lvert \sin x \lvert+C\\ \end{align*}

Formula 5:

$$\int \sec x \, dx = \ln \lvert \sec x + \tan x\lvert+C$$

Proof:

\begin{align*} \int \sec x \, dx &= \int\frac{\sec x (\sec x + \tan x) }{\sec x + \tan x} \,dx \\\\ \textup{Let}\,\, u&= \sec x + \tan x \\ du &= ( \sec x \tan x +\sec^2 x )dx\\ du &= \sec x ( \tan x +\sec x )dx\\ du &= \sec x ( \sec x + \tan x)dx\\\\ \int \sec x\, dx &= \int \frac{du}{u}\\ &= \ln \lvert u \lvert + \,C\\ &= \ln \lvert \sec x + \tan x\lvert + \,C\\ \end{align*}

Formula 6:

$$\int \sec x \, dx = \ln \left\lvert \tan \left(\frac{x}{2}+\frac{\pi}{4}\right)\right\rvert+C$$

Proof:

\begin{align*} \int \sec x \, dx &= \int\frac{1+\tan^2 \frac{x}{2}}{1-\tan^2 \frac{x}{2}} \,dx \\ &= \int\frac{\sec^2 \frac{x}{2}}{1-\tan^2 \frac{x}{2}} \,dx \\\\ \textup{Let}\,\, u&= \tan \frac{x}{2} \\ du &= \frac{1}{2} \sec^2 \frac{x}{2} dx\\ 2du &= \sec^2 \frac{x}{2} dx\\\\ &= \int \frac{2du}{1-u^2} \\ &= 2 \int \frac{1}{1-u^2} \,du \\ &= 2 \times \frac{1}{2} \ln \left\lvert \frac{1+u}{1-u}\right\rvert+C\\ &= \ln \left\lvert \frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right\rvert+C\\ &= \ln \left\lvert \frac{\tan\frac{\pi}{4}+\tan\frac{x}{2}}{1-\tan\frac{\pi}{4}\tan\frac{x}{2}}\right\rvert+C\\ &= \ln \left\lvert \tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right\rvert+C\\ &= \ln \left\lvert \tan \left(\frac{x}{2}+\frac{\pi}{4}\right)\right\rvert+C\\ \end{align*}