The antiderivatives of trigonometric functions

The sines function\(: {\displaystyle \int^x} \sin(t) \ dt \)

The \( \sin(x) \) function is defined as follows:

So by simply taking the antiderivative from each side,

The cosines function\(: \cos(x)\)

The \( \cos(x) \) function is defined as follows:

As well as above with the \(\sin(x)\) function, we directly obtain:

The tangent function\(: {\displaystyle \int^x} \tan(t) \ dt \)

The \( \tan(x) \) function is defined as follows:

From this definition, we do have:

Le us set down a new variable: \(u = \cos(t)\).

So,

The arcsines function\(: \operatorname{Arcsin}(x)\)

The \( \operatorname{Arcsin}(x) \) is the reciprocal function of the \( \sin(x) \) function , it is defined as follows:

From this definition, let us perfom an integration by parts with :

We do have:

And as a result,

The arccosines function\(: \operatorname{Arccos}(x)\)

The \( \operatorname{Arccos}(x) \) function is the reciprocal function of the \( \cos(x) \) function , it is defined as follows:

As well as above, we perform an integration by parts with:

We do have:

As a result we do have,

The arctangent function\(: {\displaystyle \int^x} \operatorname{Arctan}(t) \ dt \)

The \( \operatorname{Arctan}(x) \) function is the reciprocal function of the \( \tan(x) \) function , it is defined as follows:

From this definition, let us perfom an integration by parts with :

We do have:

And as a result,

The cosecant function\(: {\displaystyle \int^x} \csc(t) \ dt \)

1. By the secant trigonometric functions

   The \( \csc(x) \) function is defined as follows:

   $$ \forall k \in \mathbb{Z}, \ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \bigl \{ k\pi \bigr \} \Bigr], \enspace f(x) = \csc(x) = \frac{1}{\sin(x)} $$

   We firstly notice that:

   $$\csc(x) = \csc(x)\frac{\csc(x) - \cot(x)}{\csc(x) - \cot(x)}$$
   $$\csc(x) = \frac{\csc^2(x) - \csc(x)\cot(x)}{\csc(x) - \cot(x)}$$

   But,

   $$ \Biggl \{ \begin{gather*} \csc^2(x) = -\cot(x)' \\ -\csc(x)\cot(x) = \csc(x)' \end{gather*} $$

   Now we have,

   $$\csc(x) = \frac{-\cot'(x) + \csc'(x) }{\csc(x) -\cot(x)}$$
   $$\csc(x) = \frac{(\csc(x) -\cot(x))'}{\csc(x) -\cot(x)}$$

   Then, we can easily integrate it and:

   $$ \int^x \csc(t) \ dt = \int^x \frac{\bigl(\csc(t) -\cot(t) \bigr)'}{\csc(t) -\cot(t)} \ dt $$

   As a result,

   $$ \forall k \in \mathbb{Z}, \ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \bigl \{ k\pi \bigr \} \Bigr],$$
   $$ \int^x \csc(t) \ dt = \ln \left|\csc(x) -\cot(x) \right|$$

2. By applying Bioche's rules

   By applying Bioche's rules , we can put as a change of variable:

   $$ u = \tan\left(\frac{t}{2} \right) $$
   $$ \left \{ \begin{gather*} u = \tan\left(\frac{t}{2} \right) \\ du = \frac{1}{2} \left(1 + \tan^2\left(\frac{t}{2} \right) \right) dt \Longleftrightarrow du = \frac{1}{2} \left(1 + u^2 \right) \ dt \Longleftrightarrow dt = \frac{2du}{1 + u^2} \end{gather*} \right \} $$

   Then the the integral :

   $$ \int^x \csc(t) \ dt = \int^x \frac{1}{\sin(t)} \ dt $$

   becomes:

   $$ \int^x \csc(t) \ dt = \int^x \frac{1 + u^2}{2u} \ \frac{2du}{1 + u^2} $$
   $$ \int^x \csc(t) \ dt = \int^x \frac{du}{u} $$

   And finally,

   $$ \forall k \in \mathbb{Z}, \ \forall x \in \Bigl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \bigl \{ k\pi \bigr \} \Bigr],$$
   $$ \int^x \csc(t) \ dt = \ln \left| \tan \left( \frac{x}{2}\right) \right| $$

The secant function\(: {\displaystyle \int^x} \sec(t) \ dt \)

1. By the secant trigonometric functions

   The \( \sec(x) \) function is defined as follows:

   $$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \Bigl \{ \frac{\pi}{2} + k\pi \Bigr \} \biggr], \enspace f(x) = \sec(x) = \frac{1}{\cos(x)} $$

   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,

   $$ \Biggl \{ \begin{gather*} \sec^2(x) = \tan'(x) \\ \sec(x)\tan(x)= \sec'(x) \end{gather*} $$

   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:

   $$ \int^x \sec(t) \ dt = \int^x \frac{(\sec(x) + \tan(x))'}{\sec(x) + \tan(x)} \ dt $$

   And a result we do obtain,

   $$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \Bigl \{ \frac{\pi}{2} + k\pi \Bigr \} \biggr], $$
   $$ \int^x \sec(t) \ dt = \ln \left|\sec(x) + \tan(x) \right|$$

2. By applying Bioche's rules

   In the same way as above, we put the same change of variable:

   $$ u = \tan\left(\frac{t}{2} \right) $$
   $$ \left \{ \begin{gather*} u = \tan\left(\frac{t}{2} \right) \\ du = \frac{1}{2} \left(1 + \tan^2\left(\frac{t}{2} \right) \right) dt \Longleftrightarrow du = \frac{1}{2} \left(1 + u^2 \right) \ dt \Longleftrightarrow dt = \frac{2du}{1 + u^2} \end{gather*} \right \} $$

   Then the the integral :

   $$ \int^x \sec(t) \ dt = \int^x \frac{1}{\cos(t)} \ dt $$

   becomes:

   $$ \int^x \sec(t) \ dt = \int^x \frac{1 + u^2}{1 - u^2} \ \frac{2du}{1 + u^2} $$
   $$ \int^x \sec(t) \ dt = 2\int^x \frac{du}{1 - u^2} $$
   $$ \int^x \sec(t) \ dt = 2\int^x \frac{du}{(1 - u)(1 + u)} $$

   After having broken it in simple elements , we do have:

   $$ \int^x \sec(t) \ dt = 2 \times \frac{1}{2} \ln \left| \frac{1 + u}{1 - u} \right| $$

   Now, by rehabilitating the starting variable:

   $$ \int^x \sec(t) \ dt = \ln \left| \frac{1 + \tan\left(\frac{x}{2} \right)}{1 - \tan\left(\frac{x}{2} \right)} \right| $$

   But we know from the addition trigonometric formulas , that:

   $$ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta) }$$
   $$ \int^x \sec(t) \ dt = \ln \left| \frac{ \tan\left(\frac{\pi}{4} \right) + \tan\left(\frac{x}{2} \right)}{1 - \tan\left(\frac{\pi}{4} \right) \tan\left(\frac{x}{2} \right)} \right| $$

   We finally obtain that,

   $$ \forall k \in \mathbb{Z}, \enspace \forall x \in \biggl[ \mathbb{R} \hspace{0.2em} \backslash \hspace{0.2em} \Bigl \{ \frac{\pi}{2} + k\pi \Bigr \} \biggr], $$
   $$ \int^x \sec(t) \ dt = \ln \left| \tan\left( \frac{x}{2} + \frac{\pi}{4} \right) \right| $$

The cotangent function\(: {\displaystyle \int^x} \cot(t) \ dt \)

The \( \cot(x) \) function is defined as follows:

From this definition:

So,

Let us set down: \(u = \sin(t)\).

As a result we do have,

The arccosecant function\(: {\displaystyle \int^x} \operatorname{Arccsc}(t) \ dt \)

The \( \operatorname{Arccsc}(x) \) is the reciprocal function of the \( \csc(x) \) function , it is defined as follows:

From this definition, let us perfom an integration by parts with :

We do have:

We already calculated this integral above :

And finally,

The arcsecant function\(: {\displaystyle \int^x} \operatorname{Arcsec}(t) \ dt \)

The \( \operatorname{Arcsec}(x) \) is the reciprocal function of the \( \sec(x) \) function, it is defined as follows:

From this definition, by performing the same integration by parts as the \(\operatorname{Arccsc}(x)\) function above:

We directly obtain,

The arccotangent function\(: {\displaystyle \int^x} \operatorname{Arccot}(t) \ dt \)

The \( \operatorname{Arccot}(x) \) is the reciprocal function of the \( \cot(x) \) function , it is defined as follows:

From this definition, by performing the same integration by parts as the \(\operatorname{Arccsc}(x)\) function above:

We directly obtain,

The hyperbolic sines function\(: {\displaystyle \int^x} \sinh(t) \ dt \)

The \( \sinh(x) \) function is defined as follows:

As well as above with the \(\sin(x)\) function, we directly obtain:

The hyperbolic cosines function\(: {\displaystyle \int^x} \cosh(t) \ dt \)

The \( \cosh(x) \) function is defined as follows:

As well as above with the \(\sinh(x)\) function, we directly obtain:

The hyperbolic tangent function\(: {\displaystyle \int^x} \tanh(t) \ dt \)

The \( \tanh(x) \) function is defined as follows:

From this definition, we do have:

As well as above with the \(\tan(x)\) function, we set down: \(u = \cosh(t)\).

And we easily obtain,

The hyperbolic arcsines function\(: {\displaystyle \int^x} \operatorname{Argsinh}(t) \ dt \)

The \( \operatorname{Argsinh}(x) \) is the reciprocal function of the \( \sinh(x) \) function , it is defined as follows:

In addition, we can define it more explicitly by :

(\(\Longrightarrow\) see demonstration of it )

As well as above, we perform an integration by parts with:

We do have:

And as a result,

The hyperbolic arccosines function\(: {\displaystyle \int^x} \operatorname{Argcosh}(t) \ dt \)

The \( \operatorname{Argcosh}(x) \) is the reciprocal function of the \( \cosh(x) \) function , it is defined as follows:

In addition, we can define it more explicitly by :

(\(\Longrightarrow\) see demonstration of it )

As well as above, we perform an integration by parts with:

We do have:

And finally,

The hyperbolic arctangent function\(: {\displaystyle \int^x} \operatorname{Argtanh}(t) \ dt \)

The \( \operatorname{Argtanh}(x) \) is the reciprocal function of the \( \tanh(x) \) function , it is defined as follows:

In addition, we can define it more explicitly by :

(\(\Longrightarrow\) see demonstration of it )

From this definition, let us perfom an integration by parts with :

We do have:

And finally,

The hyperbolic cosecant function\(: {\displaystyle \int^x} \operatorname{csch}(t) \ dt \)

The \( \operatorname{csch}(x) \) function is defined as follows:

1. By the secant trigonometric functions

   By applying the same reasoning as above with the \(\csc(x) \) function :

   $$ \Biggl \{ \begin{gather*} \operatorname{csch}^2(x) = -\operatorname{coth}(x)' \\ -\operatorname{csch}(x)\operatorname{coth}(x) = \operatorname{csch}(x)' \end{gather*} $$

   We directly obtain that:

   $$ \forall x \in \hspace{0.04em} \mathbb{R}^*, $$
   $$ \int^x \operatorname{csch}(t) \ dt = \ln \left|\operatorname{csch}(x) -\operatorname{coth}(x) \right|$$

2. By using the change of variable \(u = e^t\)

   We now use this change of variable:

   $$ \left \{ \begin{gather*} u = e^t \\ du = e^t \ dt \Longleftrightarrow dt = \frac{du}{u} \end{gather*} \right \} $$

   Then the the integral :

   $$ \int^x \operatorname{csch}(t) \ dt = \int^x \frac{1}{\sinh(t)} \ dt = \int^x \frac{2}{e^t - e^{-t}} \ dt $$

   becomes:

   $$ \int^x \operatorname{csch}(t) \ dt = \int^x \frac{2}{u - u^{-1}} \ \frac{du}{u} $$
   $$ \int^x \operatorname{csch}(t) \ dt = 2 \int^x \frac{du}{u^2 - 1} $$

   After having broken it in simple elements , we do have:

   $$ \int^x \operatorname{csch}(t) \ dt = 2 \times \frac{1}{2} \ln \left| \frac{u + 1}{u - 1} \right| $$

   Now, by rehabilitating the starting variable:

   $$ \int^x \operatorname{csch}(t) \ dt = 2 \times \frac{1}{2} \ln \left| \frac{e^x + 1}{e^x - 1} \right| $$
   $$ \int^x \operatorname{csch}(t) \ dt = 2 \times \frac{1}{2} \ln \left| \frac{(e^{\frac{x}{2}} + e^{-\frac{x}{2}})}{e^{\frac{x}{2}}(e^{\frac{x}{2}} - e^{-\frac{x}{2}})} \right| $$
   $$ \int^x \operatorname{csch}(t) \ dt = 2 \times \frac{1}{2} \ln \left| \frac{e^{\frac{x}{2}} + e^{-\frac{x}{2}}}{e^{\frac{x}{2}} - e^{-\frac{x}{2}}} \right| $$
   $$ \int^x \operatorname{csch}(t) \ dt = \ln \left| \left( \frac{1}{\tanh \left(\frac{x}{2} \right) } \right) \right| $$

   We finally obtain that :

   $$ \forall x \in \hspace{0.04em} \mathbb{R}^*,$$
   $$ \int^x \operatorname{csch}(t) \ dt = \ln \left| \operatorname{coth}\left(\frac{x}{2} \right) \right|$$

The hyperbolic secant function\(: {\displaystyle \int^x} \operatorname{sech}(t) \ dt \)

The \( \operatorname{sech}(x) \) function is defined as follows:

1. By the secant trigonometric functions

   $$ \int^x \operatorname{sech}(t) \ dt = \int^x \frac{1}{\cosh(t)} \ dt $$
   $$ \int^x \operatorname{sech}(t) \ dt = \int^x \frac{\cosh(t)}{\cosh^2(t)} \ dt $$
   $$ \int^x \operatorname{sech}(t) \ dt = \int^x \frac{\cosh(t)}{1 + \sinh^2(t)} \ dt $$

   Let us set down the new variable: \(u = \sinh(t)\).

   Now we have:

   $$ \begin{gather*} \int^x \operatorname{sech}(t) \ dt = \int^x \frac{du}{1 + u^2} \end{gather*} $$

   $$ \text{with } \Biggl \{ \begin{gather*} u = \sinh(t) \\ du = \cosh(t) \ dt \end{gather*} $$

   And finally,

   $$ \forall x \in \mathbb{R}, $$
   $$ \int^x \operatorname{sech}(t) \ dt = \operatorname{Arctan}(\sinh(x)) $$

2. By using the change of variable \(u = e^t\)

   In the same manner as previously, we put the following change of variable:

   $$ \left \{ \begin{gather*} u = e^t \\ du = e^t \ dt \Longleftrightarrow dt = \frac{du}{u} \end{gather*} \right \} $$

   Then the the integral :

   $$ \int^x \operatorname{sech}(t) \ dt = \int^x \frac{1}{\cosh(t)} \ dt = \int^x \frac{2}{e^t + e^{-t}} \ dt $$

   becomes:

   $$ \int^x \operatorname{sech}(t) \ dt = \int^x \frac{2}{u + u^{-1}} \ \frac{du}{u} $$
   $$ \int^x \operatorname{sech}(t) \ dt = 2 \int^x \frac{du}{u^2 + 1} $$
   $$ \int^x \operatorname{sech}(t) \ dt = 2 \ \operatorname{Arctan}(u) $$

   Now, by rehabilitating the starting variable, we finally obtain that:

   $$ \forall x \in \mathbb{R}, $$
   $$ \int^x \operatorname{sech}(t) \ dt = 2 \ \operatorname{Arctan}(e^x) $$

The hyperbolic cotangent function\(: {\displaystyle \int^x} \operatorname{coth}(t) \ dt \)

The \( \operatorname{coth}(x) \) function is defined as follows:

From this definition:

As well as above, we set a new variable: \(u = \sinh(t)\).

We finally obtain,

The hyperbolic arccosecant function\(: {\displaystyle \int^x} \operatorname{Argcsch}(t) \ dt \)

The \( \operatorname{Argcsch}(x) \) is the reciprocal function of the \( \operatorname{csch}(x) \) function , it is defined as follows:

From this definition, let us perfom an integration by parts with :

We know that this integral is worth :

And finally,

The hyperbolic arcsecant function\(: {\displaystyle \int^x} \operatorname{Argsech}(t) \ dt \)

The \( \operatorname{Argsech}(x) \) is the reciprocal function of the \( \operatorname{sech}(x) \) function , it is defined as follows:

From this definition, by performing the same integration by parts as the \(\operatorname{Argcsch}(x)\) function above:

We directly obtain,

The hyperbolic arccotangent function\(: {\displaystyle \int^x} \operatorname{Argcoth}(t) \ dt \)

The \( \operatorname{Argcoth}(x) \) is the reciprocal function of the \( \operatorname{coth}(x) \) function , it is defined as follows:

From this definition, by performing the same integration by parts as the \(\operatorname{Argcsch}(x)\) function above:

We directly obtain,

Antiderivatives of trigonometric functions recap table