The basic trigonometric functions\(:sin(x), cos(x), tan(x)\)
The sines function\(: {\displaystyle \int^x} sin(t) \ dt \)
The \( sin(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = sin(x) $$
Its general antiderivative is:
$$ \forall x \in \mathbb{R}, $$
$$ \int^x sin(t) \ dt = -cos(x)$$
The cosines function\(: {\displaystyle \int^x} cos(t) \ dt \)
The \( cos(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = cos(x) $$
Its general antiderivative is:
$$ \forall x \in \mathbb{R}, $$
$$ \int^x cos(t) \ dt = sin(x)$$
The tangent function\(: {\displaystyle \int^x} tan(t) \ dt \)
The \( tan(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], \enspace f(x) = tan(x) = \frac{sin(x)}{cos(x)} $$
Its general antiderivative is:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], $$
$$ \int^x tan(t) \ dt = - ln|cos(x)| = ln|sec(x)|$$
The basic trigonometric reciprocal functions\(:arcsin(x)\), \(arccos(x)\), \( arctan(x)\)
The arcsines function\(: {\displaystyle \int^x} arcsin(t) \ dt \)
The \( arcsin(x) \) is the reciprocal function of the \( sin(x) \) function, it is defined as follows:
$$ \forall x \in [-1, \hspace{0.2em} 1], \enspace f(x) = arcsin(x) = sin^{-1}(x) $$
Its general antiderivative is:
$$ \forall x \in [-1, \hspace{0.2em} 1], $$
$$ \int^x arcsin(t) \ dt = x \ arcsin(x) + \sqrt{1-x^2}$$
The arccosines function\(: {\displaystyle \int^x} arccos(t) \ dt \)
The \( arccos(x) \) function is the reciprocal function of the \( cos(x) \) function, it is defined as follows:
$$ \forall x \in [-1, \hspace{0.2em} 1], \enspace f(x) = arccos(x) = cos^{-1}(x) $$
Its general antiderivative is:
$$ \forall x \in [-1, \hspace{0.2em} 1], $$
$$ \int^x arccos(t) \ dt = x \ arccos(x) - \sqrt{1-x^2}$$
The arctangent function\(: {\displaystyle \int^x} arctan(t) \ dt \)
The \( arctan(x) \) function is the reciprocal function of the \( tan(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = arctan(x) = tan^{-1}(x) $$
Its general antiderivative is:
$$ \forall x \in \mathbb{R}, $$
$$\int^x arctan(t) \ dt = x \ arctan(x) - \frac{1}{2} ln\left(1+x^2 \right)$$
The secant trigonometric functions: \(cosec(x), sec(x), cotan(x)\)
The cosecant function\(: {\displaystyle \int^x} cosec(t) \ dt \)
The \( cosec(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr], \enspace f(x) = cosec(x) = \frac{1}{sin(x)} $$
Its general antiderivative is:
$$ \forall k \in \mathbb{Z}, \ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr],$$
$$\int^x cosec(t) \ dt = ln \left|cosec(x) -cotan(x) \right|$$
The secant function\(: {\displaystyle \int^x} sec(t) \ dt \)
The \( sec(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], \enspace f(x) = sec(x) = \frac{1}{cos(x)} $$
Its general antiderivative is:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], $$
$$\int^x sec(t) \ dt = ln \left|sec(x) + tan(x) \right|$$
The cotangent function\(: {\displaystyle \int^x} cotan(t) \ dt \)
The \( cotan(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr] , \enspace f(x) = cotan(x) = \frac{cosec(x)}{sec(x)} = \frac{1}{tan(x)} $$
Its general antiderivative is:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ k\pi \Bigr\} \biggr] , $$
$$ \int^x cotan(t) \ dt = - ln|sin(x)| = ln|cosec(x)|$$
The secant trigonometric reciprocal functions: \(arccosec(x)\), \(arcsec(x)\), \( arccotan(x)\)
The arccosecant function\(: {\displaystyle \int^x} arccosec(t) \ dt \)
The \( arccosec(x) \) is the reciprocal function of the \( cosec(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1] \cup[1, \hspace{0.1em} +\infty[ , \enspace f(x) = arccosec(x) = cosec^{-1}(x) $$
Its general antiderivative is:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1] \cup[1, \hspace{0.1em} +\infty[ , $$
$$\int^x arccosec(t) \ dt = x \ arccosec(x) + ln \left|\sqrt{x^2-1} + |x| \right|$$
The arcsecant function\(: {\displaystyle \int^x} arcsec(t) \ dt \)
The \( arcsec(x) \) is the reciprocal function of the \( sec(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1] \cup[1, \hspace{0.1em} +\infty[ , \enspace f(x) = arcsec(x) = sec^{-1}(x) $$
Its general antiderivative is:
$$ \forall x \in \mathbb{R}, $$
$$\int^x arcsec(t) \ dt = x \ arcsec(x) - ln \left|\sqrt{x^2-1} + |x| \right| $$
The arccotangent function\(: {\displaystyle \int^x} arccotan(t) \ dt \)
The \( arccotan(x) \) is the reciprocal function of the \( cotan(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R} , \enspace f(x) = arccotan(x) = cotan^{-1}(x) $$
Its general antiderivative is:
$$ \forall x \in \mathbb{R}, $$
$$\int^x arccotan(t) \ dt = x \ arccotan(x) + \frac{1}{2} ln\left(1+x^2 \right) $$
The hyperbolic functions: \(sinh(x), cosh(x), tanh(x)\)
The hyperbolic sines function\(: {\displaystyle \int^x} sinh(t) \ dt \)
The \( sinh(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = sinh(x) = \frac{e^x - e^{-x} }{2} $$
Its general antiderivative is:
$$ \forall x \in \mathbb{R}, $$
$$ \int^x sinh(t) \ dt = cosh(x)$$
The hyperbolic cosines function\(: {\displaystyle \int^x} cosh(t) \ dt \)
The \( cosh(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = cosh(x) = \frac{e^x + e^{-x} }{2} $$
Its general antiderivative is:
$$ \forall x \in \mathbb{R}, $$
$$ \int^x cosh(t) \ dt = sinh(x)$$
The hyperbolic tangent function\(: {\displaystyle \int^x} tanh(t) \ dt \)
The \( tanh(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = tanh(x) = \frac{sinh(x)}{cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}} $$
Its general antiderivative is:
$$ \forall x \in \mathbb{R}, $$
$$ \int^x tanh(t) \ dt = ln|cosh(x)| = -ln|sech(x)|$$
The hyperbolic reciprocal functions: \( arcsinh(x)\), \(arccosh(x)\), \( arctanh(x)\)
The hyperbolic arcsines function\(: {\displaystyle \int^x} arcsinh(t) \ dt \)
The \( arcsinh(x) \) is the reciprocal function of the \( sinh(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = arcsinh(x)= sinh^{-1}(x) $$
In addition, we can define it more explicitly by :
$$ \forall x \in \mathbb{R},$$
$$ arcsinh(x) = ln \left|x + \sqrt{x^2 + 1}\right| $$
(\(\Longrightarrow\) see demonstration of it)
Its general antiderivative is:
$$ \forall x \in \mathbb{R}, $$
$$ \int^x arcsinh(t) \ dt = x \ arcsinh(x) - \sqrt{1+x^2}$$
The hyperbolic arccosines function\(: {\displaystyle \int^x} arccosh(t) \ dt \)
The \( arccosh(x) \) is the reciprocal function of the \( cosh(x) \) function, it is defined as follows:
$$ \forall x \in [1, \hspace{0.1em} +\infty[, \enspace f(x) = arccosh(x) = cosh^{-1}(x) $$
In addition, we can define it more explicitly by :
$$ \forall x \in [1, \hspace{0.1em} +\infty[, $$
$$ arccosh(x) = ln \Bigl| x + \sqrt{x^2 - 1}\Bigr| $$
(\(\Longrightarrow\) see demonstration of it)
Its general antiderivative is:
$$ \forall x \in \mathbb{R}, $$
$$ \int^x arccosh(t) \ dt = x \ arccosh(x) - \sqrt{x^2-1}$$
The hyperbolic arctangent function\(: {\displaystyle \int^x} arctanh(t) \ dt \)
The \( arcsinh(x) \) is the reciprocal function of the \( sinh(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = arcsinh(x)= sinh^{-1}(x) $$
In addition, we can define it more explicitly by :
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, $$
$$ arctanh(x) = \frac{1}{2} ln \left| \frac{1 + x}{1 - x} \right| $$
(\(\Longrightarrow\) see demonstration of it)
Its general antiderivative is:
$$ \forall x \in [-1, \hspace{0.2em} 1], $$
$$ \int^x arctanh(t) \ dt = x \ arctanh(x) + ln|1 - x^2|$$
The hyperbolic secant functions: \(cosech(x), sech(x), cotanh(x)\)
The hyperbolic cosecant function\(: {\displaystyle \int^x} cosech(t) \ dt \)
The \( cosech(x) \) function is defined as follows:
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr], \enspace f(x) = cosech(x) = \frac{1}{sinh(x)} $$
Its general antiderivative is:
$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, $$
$$\int^x cosech(t) \ dt = ln \left|cosech(x) -cotanh(x) \right|$$
The hyperbolic secant function\(: {\displaystyle \int^x} sech(t) \ dt \)
The \( sech(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = sech(x) = \frac{1}{cosh(x)} $$
Its general antiderivative is:
$$ \forall x \in \mathbb{R}, $$
$$\int^x sech(t) \ dt = arctan(sinh(x)) $$
The hyperbolic cotangent function\(: {\displaystyle \int^x} cotanh(t) \ dt \)
The \( cotanh(x) \) function is defined as follows:
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ 0 \bigr\} \Bigr], \enspace f(x) = cotanh(x) = \frac{1}{tanh(x)} $$
Its general antiderivative is:
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.1em}\backslash \left\{ 0 \right \} \Bigr], $$
$$ \int^x cotanh(t) \ dt = ln|sinh(x)| = -ln|cosech(x)|$$
The hyperbolic secant reciprocal functions: \(arccosech(x)\), \(arcsech(x)\), \( arccotanh(x)\)
The hyperbolic arccosecant function\(: {\displaystyle \int^x} arccosech(t) \ dt \)
The \( arccosech(x) \) is the reciprocal function of the \( cosech(x) \) function, it is defined as follows:
$$\forall x \in \Bigl[ \mathbb{R} \hspace{0.1em} \backslash \left \{ 0 \right \} \Bigr] , \enspace f(x) = arccosech(x) = cosech^{-1}(x) $$
Its general antiderivative is:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1] \cup[1, \hspace{0.1em} +\infty[ , $$
$$\int^x arccosech(t) \ dt = x \ arccosech(x) + ln \left|\sqrt{x^2+1} + |x| \right|$$
The hyperbolic arcsecant function\(: {\displaystyle \int^x} arcsech(t) \ dt \)
The \( arcsech(x) \) is the reciprocal function of the \( sech(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]0, \hspace{0.1em} 1] , \enspace f(x) = arcsech(x) = sech^{-1}(x) $$
Its general antiderivative is:
$$ \forall x \in \hspace{0.05em} ]0, \hspace{0.1em} 1]$$
$$\int^x arcsech(t) \ dt = x \ arcsec(x) + arcsin(x) $$
The hyperbolic arccotangent function\(: {\displaystyle \int^x} arccotanh(t) \ dt \)
The \( arccotanh(x) \) is the reciprocal function of the \( cotanh(x) \) function, it is defined as follows:
$$ \forall \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[ \hspace{0.1em} \cup \hspace{0.1em} ]1, \hspace{0.1em} +\infty[ , \enspace f(x) = arccotanh(x) =cotanh^{-1}(x) $$
Its general antiderivative is:
$$ \forall \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[ \hspace{0.1em} \cup \hspace{0.1em} ]1, \hspace{0.1em} +\infty[ , $$
$$\int^x arccotanh(t) \ dt = x \ arccotanh(x) + ln \left|1-x^2 \right| $$
Antiderivatives of trigonometric functions recap table
The \( sin(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = sin(x) $$
As we know from the derivatives of trigonometric functions that:
$$ \forall x \in \mathbb{R}, $$
$$ cos(x)' = -sin(x) $$
So by simply taking the antiderivative from each side,
$$ \int^x cos(x)' \ dt = - \int^x sin(x) \ dt $$
$$ \forall x \in \mathbb{R}, $$
$$ \int^x sin(t) \ dt = -cos(x)$$
The \( cos(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = cos(x) $$
As well as above with the \(sin(x)\) function, we directly obtain:
$$ \forall x \in \mathbb{R}, $$
$$ \int^x cos(t) \ dt = sin(x)$$
The \( tan(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], \enspace f(x) = tan(x) = \frac{sin(x)}{cos(x)} $$
From this definition, we do have:
$$\int^x tan(t) \ dt = \int^x \frac{sin(t)}{cos(t)} \ dt $$
Le us set down a new variable: \(u = cos(t)\).
$$ \begin{align*} \int^x \frac{sin(t)}{cos(t)} \ dt = \int^x -\frac{du}{u} \end{align*} $$
$$ with \enspace \Biggl \{ \begin{align*} u = cos(t) \\ du = -sin(t) \ dt \end{align*} $$
So,
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], $$
$$ \int^x tan(t) \ dt = - ln|cos(x)| = ln|sec(x)|$$
The \( arcsin(x) \) is the reciprocal function of the \( sin(x) \) function, it is defined as follows:
$$ \forall x \in [-1, \hspace{0.2em} 1], \enspace f(x) = arcsin(x) = sin^{-1}(x) $$
From this definition, let us perfom an integration by parts with :
$$ \ \Biggl \{ \begin{align*} u(t) = arcsin(t) \\ v'(t) = dt \end{align*} $$
$$ \Biggl \{ \begin{align*} u'(t) = \frac{dt}{\sqrt{1-t^2}} \\ v(t) = t \end{align*} $$
We do have:
$$\int^x arcsin(t) \ dt = \Biggl[t \ arcsin(t) \Biggr]^x -\int^x \frac{t}{\sqrt{1-t^2}} \ dt$$
$$\int^x arcsin(t) \ dt = x \ arcsin(x) - \int^x \frac{-2t}{2\sqrt{1-t^2}} \ dt$$
And as a result,
$$ \forall x \in [-1, \hspace{0.2em} 1], $$
$$ \int^x arcsin(t) \ dt = x \ arcsin(x) + \sqrt{1-x^2}$$
The \( arccos(x) \) function is the reciprocal function of the \( cos(x) \) function, it is defined as follows:
$$ \forall x \in [-1, \hspace{0.2em} 1], \enspace f(x) = arccos(x) = cos^{-1}(x) $$
As well as above, we perform an integration by parts with: :
$$ \ \Biggl \{ \begin{align*} u(t) = arccos(t) \\ v'(t) = dt \end{align*} $$
$$ \Biggl \{ \begin{align*} u'(t) = -\frac{dt}{\sqrt{1-t^2}} \\ v(t) = t \end{align*} $$
We do have:
$$\int^x arccos(t) \ dt = \Biggl[t \ arccos(t) \Biggr]^x -\int^x \frac{-t}{\sqrt{1-t^2}} \ dt$$
$$\int^x arccos(t) \ dt = x \ arccos(x) - \int^x \frac{-2t}{2\sqrt{1-t^2}} \ dt$$
As a result we do have,
$$ \forall x \in [-1, \hspace{0.2em} 1], $$
$$ \int^x arccos(t) \ dt = x \ arccos(x) - \sqrt{1-x^2}$$
The \( arctan(x) \) function is the reciprocal function of the \( tan(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = arctan(x) = tan^{-1}(x) $$
From this definition, let us perfom an integration by parts with :
$$ \ \Biggl \{ \begin{align*} u(t) = arctan(t) \\ v'(t) = dt \end{align*} $$
$$ \Biggl \{ \begin{align*} u'(t) = \frac{dt}{1+t^2} \\ v(t) = t \end{align*} $$
We do have:
$$\int^x arctan(t) \ dt = \Biggl[t \ arctan(t) \Biggr]^x -\int^x \frac{t}{1+t^2} \ dt$$
$$\int^x arctan(t) \ dt = x \ arctan(x) - \frac{1}{2} \int^x \frac{2t}{1+t^2} \ dt$$
$$\int^x arctan(t) \ dt = x \ arctan(x) - \frac{1}{2} ln\left(1+x^2 \right)$$
And as a result,
$$ \forall x \in \mathbb{R}, $$
$$ \int^x arctan(t) \ dt = x \ arctan(x) - \frac{1}{2} ln\left(1+x^2 \right)$$
The \( cosec(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr], \enspace f(x) = cosec(x) = \frac{1}{sin(x)} $$
We firstly notice that:
$$cosec(x) = cosec(x)\frac{cosec(x) - cotan(x)}{cosec(x) - cotan(x)}$$
$$cosec(x) = \frac{cosec^2(x) - cosec(x)cotan(x)}{cosec(x) - cotan(x)}$$
But,
$$ \Biggl \{ \begin{align*} cosec^2(x) = -cotan(x)' \\ -cosec(x)cotan(x) = cosec(x)' \end{align*} $$
Now we have,
$$cosec(x) = \frac{-cotan'(x) + cosec'(x) }{cosec(x) -cotan(x)}$$
$$cosec(x) = \frac{(cosec(x) -cotan(x))'}{cosec(x) -cotan(x)}$$
Then, we can easily integrate it and:
$$\int^x cosec(t) \ dt = \int^x \frac{(cosec(x) -cotan(x))'}{cosec(x) -cotan(x)} \ dt $$
As a result,
$$ \forall k \in \mathbb{Z}, \ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \bigl\{ k\pi \bigr\} \Bigr],$$
$$\int^x cosec(t) \ dt = ln \left|cosec(x) -cotan(x) \right|$$
The \( sec(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], \enspace f(x) = sec(x) = \frac{1}{cos(x)} $$
First of all, we notice that:
$$sec(x) = sec(x)\frac{sec(x) + tan(x)}{sec(x) + tan(x)}$$
$$sec(x) = \frac{sec^2(x) + sec(x)tan(x)}{sec(x) + tan(x)}$$
But,
$$ \Biggl \{ \begin{align*} sec^2(x) = tan'(x) \\ sec(x)tan(x)= sec'(x) \end{align*} $$
Therefore,
$$sec(x) = \frac{tan'(x) + sec'(x) }{sec(x) + tan(x)}$$
$$sec(x) = \frac{(sec(x) + tan(x))'}{sec(x) + tan(x)}$$
Now we can easily integrate it and:
$$\int^x sec(t) \ dt = \int^x \frac{(sec(x) + tan(x))'}{sec(x) + tan(x)} \ dt $$
Ans a result we do obtain,
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ \frac{\pi}{2} + k\pi \Bigr\} \biggr], $$
$$\int^x sec(t) \ dt = ln \left|sec(x) + tan(x) \right|$$
The \( cotan(x) \) function is defined as follows:
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ k\pi \Bigr\} \biggr] , \enspace f(x) = cotan(x) = \frac{cosec(x)}{sec(x)} = \frac{1}{tan(x)} $$
From this definition:
$$\int^x cotan(t) \ dt = \int^x \frac{cosec(t)}{sec(t)} \ dt $$
So,
$$\int^x cotan(t) \ dt = \int^x \frac{cos(t)}{sin(t)} \ dt $$
Let us set down: \(u = sin(t)\).
$$ \begin{align*} \int^x cotan(t) \ dt = \int^x \frac{du}{u} \end{align*} $$
$$ with \enspace \Biggl \{ \begin{align*} u = sin(t) \\ du = cos(t) \ dt \end{align*} $$
As a result we do have,
$$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \Bigl\{ k\pi \Bigr\} \biggr] , $$
$$ \int^x cotan(t) \ dt = - ln|sin(x)| = ln|cosec(x)|$$
The \( arccosec(x) \) is the reciprocal function of the \( cosec(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1] \cup[1, \hspace{0.1em} +\infty[ , \enspace f(x) = arccosec(x) = cosec^{-1}(x) $$
From this definition, let us perfom an integration by parts with :
$$ \ \Biggl \{ \begin{align*} u(t) = arccosec(t) \\ v'(t) = dt \end{align*} $$
$$ \left \{ \begin{align*} u'(t) = - \frac{dt}{ x^2} \times \frac{1}{ \sqrt{1 - \frac{1}{ x^2}}} \\ v(t) = t \end{align*} \right \} $$
We do have:
$$\int^x arccosec(t) \ dt = \Biggl[t \ arccosec(t) \Biggr]^x +\int^x \frac{1}{ t^2} \times \frac{t}{ \sqrt{1 - \frac{1}{ t^2}}} \ dt$$
$$\int^x arccosec(t) \ dt = \Biggl[t \ arccosec(t) \Biggr]^x +\int^x \frac{1}{ |t|^2} \times \frac{t}{ \sqrt{1 - \frac{1}{ t^2}}} \ dt$$
$$\int^x arccosec(t) \ dt = \Biggl[t \ arccosec(t) \Biggr]^x +\int^x \frac{t}{ |t|\sqrt{t^2 - 1}} \ dt$$
To manage the absolute value, we can set down:
$$ \Biggl \{ \begin{align*} w = |t| \\ dw = \frac{t}{|t|}dt \ \end{align*}$$
$$\int^x arccosec(t) \ dt = \Biggl[t \ arccosec(t) \Biggr]^x +\int^x \frac{1}{\sqrt{w^2 - 1}} \ dw$$
We already calculated this integral above:
$$\int^x arccosec(t) \ dt = \Biggl[t \ arccosec(t) \Biggr]^{x} +\Biggl[ ln \left|\sqrt{w^2-1} + w \right| \Biggr]^{|x|} $$
$$\int^x arccosec(t) \ dt = x \ arccosec(x) + ln \left|\sqrt{x^2-1} + |x| \right| $$
And finally,
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1] \cup[1, \hspace{0.1em} +\infty[ , $$
$$\int^x arccosec(t) \ dt = x \ arccosec(x) + ln \left|\sqrt{x^2-1} + |x| \right|$$
The \( arcsec(x) \) is the reciprocal function of the \( sec(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1] \cup[1, \hspace{0.1em} +\infty[ , \enspace f(x) = arcsec(x) = sec^{-1}(x) $$
From this definition, by performing the same integration by parts as the \(arccosec(x)\) function above:
$$ \ \Biggl \{ \begin{align*} u(t) = arcsec(t) \\ v'(t) = dt \end{align*} $$
$$ \left \{ \begin{align*} u'(t) = \frac{dt}{ x^2} \times \frac{1}{ \sqrt{1 - \frac{1}{ x^2}}} \\ v(t) = t \end{align*} \right \} $$
We directly obtain,
$$ \forall x \in \mathbb{R}, $$
$$\int^x arcsec(t) \ dt = x \ arcsec(x) - ln \left|\sqrt{x^2-1} + |x| \right| $$
The \( arccotan(x) \) is the reciprocal function of the \( cotan(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R} , \enspace f(x) = arccotan(x) = cotan^{-1}(x) $$
From this definition, by performing the same integration by parts as the \(arccosec(x)\) function above:
$$ \ \Biggl \{ \begin{align*} u(t) = arccotan(t) \\ v'(t) = dt \end{align*} $$
$$ \left \{ \begin{align*} u'(t) = -\frac{dt}{ 1 + x^2} \\ v(t) = t \end{align*} \right \} $$
We directly obtain,
$$ \forall x \in \mathbb{R}, $$
$$\int^x arccotan(t) \ dt = x \ arccotan(x) + \frac{1}{2} ln\left(1+x^2 \right) $$
The \( sinh(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = sinh(x) = \frac{e^x - e^{-x} }{2} $$
As well as above with the \(sin(x)\) function, we directly obtain:
$$ \forall x \in \mathbb{R}, $$
$$ \int^x sinh(t) \ dt = cosh(x)$$
The \( cosh(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = cosh(x) = \frac{e^x + e^{-x} }{2} $$
As well as above with the \(sinh(x)\) function, we directly obtain:
$$ \forall x \in \mathbb{R}, $$
$$ \int^x cosh(t) \ dt = sinh(x)$$
The \( tanh(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = tanh(x) = \frac{sinh(x)}{cosh(x)} = \frac{e^x - e^{-x}}{e^x + e^{-x}} $$
From this definition, we do have:
$$\int^x tanh(t) \ dt = \int^x \frac{sinh(t)}{cosh(t)} \ dt $$
As well as above with the \(tan(x)\) function, we set down: \(u = cosh(t)\).
And we easily obtain,
$$ \forall x \in \mathbb{R}, $$
$$ \int^x tanh(t) \ dt = ln|cosh(x)| = -ln|sech(x)|$$
The \( arcsinh(x) \) is the reciprocal function of the \( sinh(x) \) function, it is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = arcsinh(x)= sinh^{-1}(x) $$
In addition, we can define it more explicitly by :
$$ \forall x \in \mathbb{R}, $$
$$ arcsinh(x) = ln \left|x + \sqrt{x^2 + 1}\right| $$
(\(\Longrightarrow\) see demonstration of it)
As well as above, we perform an integration by parts with:
$$ \ \Biggl \{ \begin{align*} u(t) = arcsinh(t) \\ v'(t) = dt \end{align*} $$
$$ \Biggl \{ \begin{align*} u'(t) = \frac{dt}{\sqrt{1+t^2}} \\ v(t) = t \end{align*} $$
We do have:
$$\int^x arcsinh(t) \ dt = \Biggl[t \ arcsinh(t) \Biggr]^x -\int^x \frac{t}{\sqrt{1+t^2}} \ dt$$
$$\int^x arcsinh(t) \ dt = x \ arcsinh(x) - \int^x \frac{2t}{2\sqrt{1+t^2}} \ dt$$
And as a result,
$$ \forall x \in \mathbb{R}, $$
$$ \int^x arcsinh(t) \ dt = x \ arcsinh(x) - \sqrt{1+x^2}$$
The \( arccosh(x) \) is the reciprocal function of the \( cosh(x) \) function, it is defined as follows:
$$ \forall x \in [1, \hspace{0.1em} +\infty[, \enspace f(x) = arccosh(x) = cosh^{-1}(x) $$
In addition, we can define it more explicitly by :
$$ \forall x \in [1, \hspace{0.1em} +\infty[, $$
$$ arccosh(x) = ln \Bigl| x + \sqrt{x^2 - 1}\Bigr| $$
(\(\Longrightarrow\) see demonstration of it)
As well as above, we perform an integration by parts with:
$$ \ \Biggl \{ \begin{align*} u(t) = arccosh(t) \\ v'(t) = dt \end{align*} $$
$$ \Biggl \{ \begin{align*} u'(t) = \frac{dt}{\sqrt{t^2 - 1}} \\ v(t) = t \end{align*} $$
We do have:
$$\int^x arccosh(t) \ dt = \Biggl[t \ arccosh(t) \Biggr]^x -\int^x \frac{t}{\sqrt{t^2 - 1}} \ dt$$
$$\int^x arccosh(t) \ dt = x \ arccosh(x) - \int^x \frac{2t}{2\sqrt{t^2-1}} \ dt$$
And finally,
$$ \forall x \in \mathbb{R}, $$
$$ \int^x arccosh(t) \ dt = x \ arccosh(x) - \sqrt{x^2-1}$$
The \( arctanh(x) \) is the reciprocal function of the \( tanh(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, \enspace f(x) = arctanh(x) = tanh^{-1}(x) $$
In addition, we can define it more explicitly by :
$$ \forall x \in \hspace{0.05em} ]-1, \hspace{0.1em} 1[, \forall x \in [1, \hspace{0.1em} +\infty[, $$
$$ \ arctanh(x) = \frac{1}{2} ln \left| \frac{1 + x}{1 - x} \right| $$
(\(\Longrightarrow\) see demonstration of it)
From this definition, let us perfom an integration by parts with :
$$ \ \Biggl \{ \begin{align*} u(t) = arctanh(t) \\ v'(t) = dt \end{align*} $$
$$ \Biggl \{ \begin{align*} u'(t) = \frac{dt}{1-x^2} \\ v(t) = t \end{align*} $$
We do have:
$$\int^x arctanh(t) \ dt = \Biggl[t \ arctanh(t) \Biggr]^x -\int^x \frac{t}{1-t^2} \ dt$$
$$\int^x arctanh(t) \ dt = x \ arctanh(x) + \frac{1}{2}\int^x \frac{-2t}{1-t^2} \ dt$$
And finally,
$$ \forall x \in [-1, \hspace{0.2em} 1], $$
$$ \int^x arctanh(t) \ dt = x \ arctanh(x) + ln|1 + x^2|$$
The \( cosech(x) \) function is defined as follows:
$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, \enspace f(x) = cosech(x) = \frac{1}{sinh(x)} $$
By applying the same reasoning as above with the \(cosec(x) \) function :
$$ \Biggl \{ \begin{align*} cosech^2(x) = -cotanh(x)' \\ -cosech(x)cotanh(x) = cosech(x)' \end{align*} $$
We directly obtain that:
$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, $$
$$\int^x cosech(t) \ dt = ln \left|cosech(x) -cotanh(x) \right|$$
The \( sech(x) \) function is defined as follows:
$$ \forall x \in \mathbb{R}, \enspace f(x) = sech(x) = \frac{1}{cosh(x)} $$
So,
$$ \int^x sech(t) \ dt = \int^x \frac{1}{cosh(t)} \ dt $$
$$ \int^x sech(t) \ dt = \int^x \frac{cosh(t)}{cosh^2(t)} \ dt $$
$$ \int^x sech(t) \ dt = \int^x \frac{cosh(t)}{1 + sinh^2(t)} \ dt $$
Let us set down the new variable: \(u = sinh(t)\).
Now we have:
$$ \begin{align*} \int^x sech(t) \ dt = \int^x \frac{du}{1 + u^2} \end{align*} $$
$$ with \enspace \Biggl \{ \begin{align*} u = sinh(t) \\ du = cosh(t) \ dt \end{align*} $$
And finally,
$$ \forall x \in \mathbb{R}, $$
$$\int^x sech(t) \ dt = arctan(sinh(x)) $$
The \( cotanh(x) \) function is defined as follows:
$$ \forall x \in \hspace{0.05em} \mathbb{R}^*, \enspace f(x) = cotanh(x) = \frac{1}{tanh(x)} $$
From this definition:
$$\int^x cotanh(t) \ dt = \int^x \frac{cosh(t)}{sinh(t)} \ dt $$
As well as above, we set a new variable: \(u = sinh(t)\).
$$ \begin{align*} \int^x cotanh(t) \ dt = \int^x \frac{du}{u} \end{align*} $$
$$ with \enspace \Biggl \{ \begin{align*} u = sinh(t) \\ du = cosh(t) \ dt \end{align*} $$
We finally obtain,
$$ \forall x \in \Bigl[ \mathbb{R} \hspace{0.1em}\backslash \left\{ 0 \right \} \Bigr], $$
$$ \int^x cotanh(t) \ dt = ln|sinh(x)| = -ln|cosech(x)|$$
The \( arccosech(x) \) is the reciprocal function of the \( cosech(x) \) function, it is defined as follows:
$$\forall x \in \Bigl[ \mathbb{R} \hspace{0.1em} \backslash \left \{ 0 \right \} \Bigr] , \enspace f(x) = arccosech(x) = cosech^{-1}(x) $$
From this definition, let us perfom an integration by parts with :
$$ \ \Biggl \{ \begin{align*} u(t) = arccosech(t) \\ v'(t) = dt \end{align*} $$
$$ \left \{ \begin{align*} u'(t) = - \frac{dt}{ x^2} \times \frac{1}{ \sqrt{1 + \frac{1}{ x^2}}} \\ v(t) = t \end{align*} \right \} $$
$$\int^x arccosech(t) \ dt = \Biggl[t \ arccosech(t) \Biggr]^x +\int^x \frac{1}{ t^2} \times \frac{t}{ \sqrt{1 + \frac{1}{ t^2}}} \ dt$$
$$\int^x arccosech(t) \ dt = \Biggl[t \ arccosech(t) \Biggr]^x +\int^x \frac{1}{ |t|^2} \times \frac{t}{ \sqrt{1 + \frac{1}{ x^2}}} \ dt$$
$$\int^x arccosech(t) \ dt = \Biggl[t \ arccosech(t) \Biggr]^x +\int^x \frac{t}{ |t|\sqrt{t^2 + 1}} \ dt$$
As well as above, we set down:
$$ \Biggl \{ \begin{align*} w = |t| \\ dw = \frac{t}{|t|}dt \ \end{align*}$$
$$\int^x arccosech(t) \ dt = \Biggl[t \ arccosech(t) \Biggr]^x +\int^x \frac{1}{\sqrt{w^2 + 1}} \ dw$$
We already calculated this integral above:
$$\int^x arccosech(t) \ dt = \Biggl[t \ arccosech(t) \Biggr]^x +\Biggl[ ln \left|\sqrt{w^2+1} + w \right| \Biggr]^{|x|} $$
$$\int^x arccosech(t) \ dt = x \ arccosech(x) + ln \left|\sqrt{x^2+1} + |x| \right| $$
And finally,
$$ \forall x \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1] \cup[1, \hspace{0.1em} +\infty[ , $$
$$\int^x arccosech(t) \ dt = x \ arccosech(x) + ln \left|\sqrt{x^2+1} + |x| \right|$$
The \( arcsech(x) \) is the reciprocal function of the \( sech(x) \) function, it is defined as follows:
$$ \forall x \in \hspace{0.05em} ]0, \hspace{0.1em} 1] , \enspace f(x) = arcsech(x) = sech^{-1}(x) $$
From this definition, by performing the same integration by parts as the \(arccosech(x)\) function above:
$$ \ \Biggl \{ \begin{align*} u(t) = arcsech(t) \\ v'(t) = dt \end{align*} $$
$$ \left \{ \begin{align*} u'(t) = - \frac{dt}{ x^2} \times \frac{1}{ \sqrt{\frac{1}{ x^2} - 1}} \\ v(t) = t \end{align*} \right \} $$
We directly obtain,
$$ \forall x \in \hspace{0.05em} ]0, \hspace{0.1em} 1] , \enspace f(x) = arcsech(x) = sech^{-1}(x) $$
$$\int^x arcsech(t) \ dt = x \ arcsec(x) + arcsin(x) $$
The \( arccotanh(x) \) is the reciprocal function of the \( cotanh(x) \) function, it is defined as follows:
$$ \forall \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[ \hspace{0.1em} \cup \hspace{0.1em} ]1, \hspace{0.1em} +\infty[ , \enspace f(x) = arccotanh(x) =cotanh^{-1}(x) $$
From this definition, by performing the same integration by parts as the \(arccosech(x)\) function above:
$$ \ \Biggl \{ \begin{align*} u(t) = arccotanh(t) \\ v'(t) = dt \end{align*} $$
$$ \left \{ \begin{align*} u'(t) = \frac{dt}{1-x^2} \\ v(t) = t \end{align*} \right \} $$
We directly obtain,
$$ \forall \in \hspace{0.05em} ]-\infty, \hspace{0.1em} -1[ \hspace{0.1em} \cup \hspace{0.1em} ]1, \hspace{0.1em} +\infty[ , $$
$$\int^x arccotanh(t) \ dt = x \ arccotanh(x) + ln \left|1-x^2 \right| $$
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