Indefinite Integrals of $trig(x)$
Index
\[\begin{aligned}
& \int \sin(x) \,\mathrm{d}x = -\cos(x) + C \\
& \int \cos(x) \,\mathrm{d}x = \sin(x) + C \\
& \int \tan(x) \,\mathrm{d}x = -\ln(\vert \cos(x) \vert) + C = \ln(\vert \sec(x) \vert) + C \\
& \int \cot(x) \,\mathrm{d}x = \ln(\vert \sin(x) \vert) + C = -\ln(\vert \csc(x) \vert) + C \\
& \int \sec(x) \,\mathrm{d}x = \ln(\vert \sec(x)+\tan(x) \vert) + C = - \ln(\vert \sec(x)-\tan(x) \vert) + C \\
& \int \csc(x) \,\mathrm{d}x = -\ln(\vert \csc(x)+\cot(x) \vert) + C = \ln(\vert \csc(x)-\cot(x) \vert) + C
\end{aligned}\]
sin
\[\begin{aligned}
&\dfrac{\mathrm{d}}{\mathrm{d}x}\cos(x) = -\sin(x) \\
&\therefore \dfrac{\mathrm{d}}{\mathrm{d}x}(-\cos(x)) = \sin(x) \\
&\therefore \int \sin(x) \,\mathrm{d}x = -\cos(x) + C
\end{aligned}\]
cos
\[\begin{aligned}
&\dfrac{\mathrm{d}}{\mathrm{d}x}\sin(x) = \cos(x) \\
&\therefore \int \cos(x) \,\mathrm{d}x = \sin(x) + C
\end{aligned}\]
tan
\[\begin{aligned}
\int \tan(x) \,\mathrm{d}x &= \int \dfrac{\sin(x)}{\cos(x)} \,\mathrm{d}x \\
&= \int \dfrac{1}{\cos(x)} \,\mathrm{d}(-\cos(x)) \\
&= -\int \dfrac{1}{\cos(x)} \,\mathrm{d}(\cos(x)) \\
&= -\ln(\vert \cos(x) \vert) + C \\
&\overset{\text{or}}{=} \ln(\vert \sec(x) \vert) + C
\end{aligned}\]
cot
\[\begin{aligned}
\int \cot(x) \,\mathrm{d}x &= \int \dfrac{\cos(x)}{\sin(x)} \,\mathrm{d}x \\
&= \int \dfrac{1}{\sin(x)} \,\mathrm{d}(\sin(x)) \\
&= \ln(\vert \sin(x) \vert) + C \\
&\overset{\text{or}}{=} -\ln(\vert \csc(x) \vert) + C
\end{aligned}\]
sec
\[\begin{aligned}
\dfrac{\mathrm{d}}{\mathrm{d}x} \sec(x) &= \sec(x)\tan(x) \\
\dfrac{\mathrm{d}}{\mathrm{d}x} \tan(x) &= \sec^2(x) \\
\therefore
\dfrac{\mathrm{d}}{\mathrm{d}x} (\sec(x) + \tan(x)) &= \sec(x)(\sec(x) + \tan(x))\\
\\
\therefore
\int \sec(x) \,\mathrm{d}x &= \int \sec(x) \dfrac{\sec(x) + \tan(x)}{\sec(x) + \tan(x)} \,\mathrm{d}x \\
&= \int \dfrac{\sec(x)(\sec(x) + \tan(x))}{\sec(x) + \tan(x)} \,\mathrm{d}x \\
&= \int \dfrac{1}{\sec(x) + \tan(x)} \, \mathrm{d}(\sec(x) + \tan(x))\\
&\overset{u=\sec(x) + \tan(x)}{=} \int \dfrac{1}{u} \, \mathrm{d}u\\
&= \ln(\vert u \vert) + C \\
&= \ln(\vert \sec(x) + \tan(x) \vert) + C \\
\\
\because \dfrac{1}{\sec(x) + \tan(x)} &= \sec(x) - \tan(x) \\
\\
\therefore \int \sec(x) \,\mathrm{d}x &= \ln(\vert \sec(x) + \tan(x) \vert) + C = -\ln(\vert \sec(x) - \tan(x) \vert) + C
\end{aligned}\]
PS: $\dfrac{1}{\sec(x) + \tan(x)} = \sec(x) - \tan(x)$ is proved here.
csc
\[\begin{aligned}
\dfrac{\mathrm{d}}{\mathrm{d}x} \csc(x) &= -\csc(x)\cot(x) \\
\dfrac{\mathrm{d}}{\mathrm{d}x} \cot(x) &= -\csc^2(x) \\
\therefore
\dfrac{\mathrm{d}}{\mathrm{d}x} (\csc(x) + \cot(x)) &= -\csc(x)(\csc(x) + \cot(x))\\
\\
\therefore
\int \csc(x) \,\mathrm{d}x &= \int \csc(x) \dfrac{\csc(x) + \cot(x)}{\csc(x) + \cot(x)} \,\mathrm{d}x \\
&= \int \dfrac{\csc(x)(\csc(x) + \cot(x))}{\csc(x) + \cot(x)} \,\mathrm{d}x \\
&= -\int \dfrac{1}{\csc(x) + \cot(x)} \, \mathrm{d}(\csc(x) + \cot(x))\\
&\overset{u=\csc(x) + \cot(x)}{=} -\int \dfrac{1}{u} \, \mathrm{d}u\\
&= -\ln(\vert u \vert) + C \\
&= -\ln(\vert \csc(x) + \cot(x) \vert) + C \\
\\
\because \dfrac{1}{\csc(x) + \cot(x)} &= \csc(x) - \cot(x) \\
\\
\therefore \int \csc(x) \,\mathrm{d}x &= -\ln(\vert \csc(x) + \cot(x) \vert) + C = \ln(\vert \csc(x)-\cot(x) \vert) + C
\end{aligned}\]
PS: $\dfrac{1}{\csc(x) + \cot(x)} = \csc(x) - \cot(x)$ is proved here.