\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*}