\begin{array}{c}
\int \sin x \mathrm{d}x=-\cos x + C\\
\int \cos x \mathrm{d}x=\sin x + C\\
\int \csc x \mathrm{d}x=\ln \left | \csc x-\cot x \right| + C\\
\int \sec x \mathrm{d}x=\ln \left | \sec x+\tan x \right| + C\\
\int \tan x \mathrm{d}x=-\ln \left | \cos x \right | + C\\
\int \cot x \mathrm{d}x=\ln \left | \sin x \right | + C\\
\int {\sin}^{2} x \mathrm{d}x=\int \frac{1-\cos 2x}{2} \mathrm{d}x = \frac{1}{2}x-\frac{1}{4}\sin 2x + C \\
\int {\cos}^{2} x \mathrm{d}x=\int \frac{1+\cos 2x}{2} \mathrm{d}x = \frac{1}{2}x+\frac{1}{4}\sin 2x + C \\
\int {\csc}^{2} x \mathrm{d}x=-\cot x + C \\
\int {\sec}^{2} x \mathrm{d}x=\tan x + C \\
\int {\tan}^{2} x \mathrm{d}x = \int \frac{1-{\cos}^{2}x}{{\cos}^{2}x}\mathrm{d}x = \tan x - x + C \\
\int {\cot}^{2} x \mathrm{d}x = \int \frac{1-{\sin}^{2}x}{{\sin}^{2}x}\mathrm{d}x= -\cot x - x + C \\
\int \sec x \tan x \mathrm{d}x = \sec x + C \\
\int \csc x \cot x \mathrm{d}x = -\csc x + C \\
\left ( \csc x \right )' = {\left ( \frac{1}{\sin x} \right)}' = \frac{-\cos x}{{\sin}^{2}x} = - \csc x \cot x \\
\left ( \sec x \right )' = {\left ( \frac{1}{\cos x} \right)}' = \frac{-\sin x}{{\cos}^{2}x} = - \sec x \tan x \\
\end{array}
