Antiderivative of Inverse Trigonometric Functions

Antiderivative of the Arcsine function

Define $y = \arcsin v$ . Then, $v = \sin y$ . From this, it follows that $dv = \cos y \cdot dy$ .

In order to calculate the antiderivative for the arcsine function, change the expression in terms of the variable $v$ to its equivalent in terms of $y$ in the integrand as follows:

$\begin{equation*} \int \arcsin v \cdot dv = \int y\cdot \cos y\cdot dy \end{equation*}$

Now, define $u = y$ , so that $du = dy$ and also let $dw = \cos y\cdot dy$ , and from this, $w = \sin y$ . With this definitions at hand
apply the technique of integration by parts to obtain:

$\begin{equation*} \int y\cdot \cos y\cdot dy = y\,\sin y - \int \sin y\cdot dy = y\,\sin y - \cos y + \hat{C} \end{equation*}$

Since by definition, $v = \sin y$ , and also, $\sin^2 y + \cos^2 y = 1$ , it follows that $\cos^2y = 1 - v^2$ , therefore,

$\begin{equation*} \int \arcsin v \cdot dv = v\cdot \arcsin v + \sqrt{1 - v^2} + C \end{equation*}$

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Antiderivative of Inverse Trigonometric Functions

The formulas to compute antiderivatives of inverse trigonometric functiosn are deduced.

Antiderivative of the Arcsine function