$$
\definecolor{comment}{RGB}{203,23,206}
$$
$$
\newcommand{\norm}[1]{\left\Vert#1\right\Vert}
\newcommand{\abs}[1]{\left\vert#1\right\vert}
\newcommand{\set}[1]{\left\{#1\right\}}
\newcommand{\Comp}{\mathbb C}
\newcommand{\Real}{\mathbb R}
\newcommand{\Nat}{\mathbb N}
\newcommand{\of}[1]{\left ( #1 \right ) }
\newcommand{\To}{\rightarrow}
\newcommand{\eps}{\varepsilon}
$$
Integral Calculus lecture notes: The arcsin function
Eugene Kritchevski, Vanier College, last updated on March 5, 2016
Definition of arcsin (secondary 5 or calculus 1)
Definition
The function $\arcsin$ has domain $[-1,1]$ and range $[-\pi/2,\pi/2]$, and is defined by
$$\theta=\arcsin(y)\Leftrightarrow y=\sin(\theta) \qquad -1\leq y\leq 1, \qquad -\pi/2\leq \theta\leq \pi/2$$
Some examples: $\sin(\pi/6)=1/2$ and $\arcsin(1/2)=\pi/6$, $\sin(\pi/2)=1$ and $\arcsin(1)=\pi/2$, $\sin(-\pi/2)=-1$ and $\arcsin(-1)=-\pi/2$, $\sin(0)=0$ and $\arcsin(0)=0$
Caution: Since $\arcsin$ is the inverse function of $\sin$, sometimes $\arcsin(y)$ is denoted by $\sin^{-1}(y)$, which DOES NOT MEAN $(\sin(y))^{-1}$.
In these notes, we use the variable $y$, because $y=\sin(\theta)$ represents the $y$ coordinate of the point representing $\theta$ on the unit circle.
Derivative of arcsin (calculus 1 or calculus 2)
Differenting $y=\sin(\theta)$ gives
$$dy=\cos(\theta)\,d\theta.$$
Since $\theta$ is in $[-\pi/2,\pi/2]$, we have $\cos(\theta) \geq 0$ and we can express $\cos(\theta)$ in terms of $y$ as follows
$$\cos(\theta)=+\sqrt{1-\sin^2(\theta)}=\sqrt{1-y^2}.$$
Then
$$dy=\sqrt{1-y^2}\,d\theta$$
and
$$d\theta=\frac{1}{\sqrt{1-y^2}}\,dy$$
which means that
$$\frac{d}{dy}\arcsin(y)=\frac{d\theta}{dy}=\frac{1}{\sqrt{1-y^2}}.$$
Note that at the endpoints $y=\pm 1$ of the domain $[-1,1]$, the expression $1/\sqrt{1-y^2}$ is undefined (the denominator is zero) and therefore $\arcsin'(\pm 1)$ does not exist in the usual sense. Actually, the graph of $\theta=\arcsin(y)$ has a vertical tangent line at $(-1,-\pi/2)$ and at $(1,\pi/2)$.
$$\arcsin'(y)=\frac{1}{\sqrt{1-y^2}}\qquad (-1 < y < 1)$$
Antiderivative of arcsin (calculus 2)
The indefinte integral
$$I=\int \arcsin(y)\,dy$$
can be evaluated using the method of integration by parts . With $u=\arcsin(y)$ and $dv=dy$ we have $du=\frac{1}{\sqrt{1-y^2}}\,dy$ and $v=y$ and so
$$I=\int \arcsin(y)\,dy=\int u\,dv=uv-\int v\,du=y\arcsin(y)-\underbrace{\int y \frac{1}{\sqrt{1-y^2}}\,dy}_J$$
To find the $J$ integral, we use the substitution $w=1-y^2$. Then $dw=-2ydy$ and $y\,dy=(-1/2)\,dw$ and so (ignoring the constant of integration at this intermediate step)
$$J=\int y \frac{1}{\sqrt{1-y^2}}\,dy=\int \frac{1}{\sqrt{w}}(-1/2)\,dw=\int w^{-1/2}(-1/2)\,dw=\frac{w^{1/2}}{1/2}(-1/2)=-w^{1/2}=-\sqrt{1-y^2}.$$
Therefore
$$I=y\arcsin(y)-J=y\arcsin(y)-(-\sqrt{1-y^2})=y\arcsin(y)+\sqrt{1-y^2} + C.$$
Conclusion:
$$\int \arcsin(y)\,dy=y\arcsin(y)+\sqrt{1-y^2} + C\qquad (-1 < y < 1)$$
Related trigonometric substitutions (calculus 2)
The $sin$ and $arcsin$ functions can sometimes help us evaluate integrals involving the expression $\sqrt{1-y^2}$ or, more generally, the expression $\sqrt{r^2-y^2}$.
Example 1: Evaluate the indefinite integral $$I=\int \sqrt{1-y^2}\,dy\qquad (-1 < y < 1).$$
Solution: We use the substitution $y=\sin(\theta)$, where $-\pi/2 < \theta < \pi/2$. The inverse substituion is $\theta=\arcsin(y)$, where $-1 < y < 1$. Then $$dy=\cos(\theta)\,d\theta,\qquad \sqrt{1-y^2}=\cos(\theta),$$ and
$$
\begin{equation*}
\begin{split}
I&=\int \sqrt{1-y^2}\,dy \\
&=\int \cos(\theta)\cos(\theta)\,d\theta \\
&=\int \frac{1}{2}\of{1+\cos(2\theta)}\,d\theta \qquad &\color{comment}\textrm{using the half-angle formula} \cos^2(\theta)=\frac{1}{2}\of{1+\cos(2\theta)}\\
&= \frac{1}{2}\of{\theta+\frac{1}{2}\sin(2\theta)} +C \\
&= \frac{1}{2}\of{\theta+\sin(\theta)\cos(\theta)} +C\qquad &\color{comment}\textrm{using } \sin(2\theta)=2\sin(\theta)\cos(\theta)\\
&= \frac{1}{2}\of{\arcsin(y)+y\sqrt{1-y^2}} +C.\\
\end{split}
\end{equation*}
$$
Example 2: Evaluate the definite integral
$$I(b)=\int_0^b \sqrt{1-y^2}\,dy\qquad (0 < b < 1)$$
and the limit
$$\lim_{b\to 1^{-}}I(b).$$
Give a geometrical interpretaion of the calculations.
Solution: Using the result of Example 1 and FTC2, we have
$$I(b)=\frac{1}{2}\of{\arcsin(y)+y\sqrt{1-y^2}}\Big|_{0}^{b}=\frac{1}{2}\of{\arcsin(b)+b\sqrt{1-b^2}}$$
and then
$$\lim_{b\to 1^{-}}I(b)=\frac{1}{2}\of{\arcsin(1)+1\sqrt{1-1^2}}=\frac{1}{2}\arcsin(1)=\frac{1}{2}\frac{\pi}{2}=\frac{\pi}{4}.$$
The integral $I(b)$ computes the area of the region in the $(x,y)$-plane given by
$$\set{(x,y): 0\leq y \leq b \textrm{ and } 0\leq x \leq \sqrt{1-y^2}}.$$
The region can be decomposed into a circular sector and a right triangle as in the figure below.
The area of the sector is $\frac{1}{2}\arcsin(b)$ and the area of the triangle is $\frac{1}{2}b\sqrt{1-b^2}$. As $b\to 1^{-}$, $I(b)$ approaches the area of the quarter of the unit circle, which is $\pi/4$:
$$\lim_{b\to 1^{-}}I(b)=\int_{0}^{1}\sqrt{1-y^2}\,dy=\frac{\pi}{4}$$
Example 3: Given $r>0$, evaluate the definite integral
$$I=\int_{-r}^{r} \of{\sqrt{r^2-y^2}}^3\,dy.$$
Solution: Since
$$\sqrt{r^2-y^2}=\sqrt{r^2\of{1-\frac{y^2}{r^2}}}=r\sqrt{1-\of{\frac{y}{r}}^2},$$
we make the substitution
$$\frac{y}{r}=\sin(\theta),\qquad -r \leq y \leq r,\qquad -\pi/2 \leq \theta \leq \pi/2$$
Then
$$y= r\sin(\theta)$$
$$dy=r\cos(\theta)\, d\theta$$
$$\sqrt{r^2-y^2}=r\cos(\theta)$$
$$y=-r \Leftrightarrow \theta =-\pi/2$$
$$y=r \Leftrightarrow \theta =\pi/2$$
and
$$
\begin{equation*}
\begin{split}
I&=\int_{-r}^{r} \of{\sqrt{r^2-y^2}}^3\,dy\\
&=\int_{-\pi/2}^{\pi/2}(r\cos(\theta))^3 r\cos(\theta)\,d\theta\\
&=r^4\int_{-\pi/2}^{\pi/2}\cos^4(\theta)\,d\theta\\
&=r^4\int_{-\pi/2}^{\pi/2}\of{\cos^2(\theta)}^2\,d\theta\\
&=r^4\int_{-\pi/2}^{\pi/2}\of{\frac{1}{2}\of{1+\cos(2\theta)}}^2\,d\theta\\
&=\frac{r^4}{4}\int_{-\pi/2}^{\pi/2}\of{1+2\cos(2\theta) +\cos^2(2\theta)}\,d\theta\\
&=\frac{r^4}{4}\int_{-\pi/2}^{\pi/2}\of{1+2\cos(2\theta) +\frac{1}{2}\of{1+\cos(4\theta)}}\,d\theta\\
&=\frac{r^4}{4}\int_{-\pi/2}^{\pi/2}\of{\frac{3}{2}+2\cos(2\theta) +\frac{1}{2}\cos(4\theta)}\,d\theta\\
&=\frac{r^4}{4}\of{\frac{3}{2}\theta+\sin(2\theta) +\frac{1}{8}\sin(4\theta)}\Big|_{-\pi/2}^{\pi/2}\\
&=\frac{3\pi r^4}{8}
\end{split}
\end{equation*}
$$
(Optional remark for Honours NYB or for Calculus 3) Geometrical interpretaion: The ball of radius $r$ in the four-dimensional space $\Real^4$ is defined by
$$B_4(r)=\set{(x_1,x_2,x_3,x_4): x_1^2+x_2^2+x_3^2+x_4^2 \leq r^2}.$$ Given $t$ in $[-r,r]$, the cross-section $S_t$ of the ball perpendicular to the $x_4$-axis
is obtained by freezing the $x_4$ coordinate to be $x_4=t$, so
$$S_t=\set{x_1^2+x_2^2+x_3^2+t^2=r^2}=\set{x_1^2+x_2^2+x_3^2=r^2-t^2},$$
which is a ball in the three dimemensional space $\Real^3$ of radius $\sqrt{r^2-t^2}$. The (three dimensional) volume of $S_t$ is then
$$vol(S_t)=\frac{4}{3}\pi (radius)^3= \frac{4}{3}\pi\of{\sqrt{r^2-t^2}}^3.$$
The (four-dimensional) volume of $B_4(r)$ is computed by integrating the (three-dimensional) cross-sectional volume with respect to $t$:
$$vol(B_4(r))=\int_{-r}^{r}vol(S_t)\,dt=\int_{-r}^{r}\frac{4}{3}\pi\of{\sqrt{r^2-t^2}}^3\,dt=\frac{4}{3}\pi\int_{-r}^{r}\of{\sqrt{r^2-t^2}}^3\,dt=\frac{4}{3}\pi \frac{3\pi r^4}{8}=\frac{1}{2}\pi^2 r^2.$$
Example 4: Evaluate the indefinite integral
$$I=\int \frac{y^3}{\sqrt{25-y^2}}\,dy.$$
Solution: We use the substitution
$$y=5\sin(\theta), \qquad -5 \leq y \leq 5, -\pi/2 \leq y \leq \pi/2$$
Then
$$dy=5\cos(\theta)\,d\theta$$
and
$$\sqrt{25-y^2}=5\cos(\theta)$$
and
$$
\begin{equation*}
\begin{split}
I&=\int \frac{y^3}{\sqrt{25-y^2}}\,dy\\
&=\int \frac{(5\sin(\theta))^3}{5\cos(\theta)}\,5\cos(\theta)\,d\theta\\
&=125\int \sin^3(\theta)\,d\theta \\
&=125\int \sin^2(\theta)\sin(\theta)\,d\theta \\
&=125\int (1-\cos^2(\theta))\sin(\theta)\,d\theta \\
&=125\int (1-u^2) (-du) \qquad &\color{comment}\textrm{using the substitution } u=\cos(\theta), \,du=-\sin(\theta)\,d\theta\\
&=125 \of{-u+\frac{1}{3}u^3} +C \\
&=125 \of{-\cos(\theta)+\frac{1}{3}\cos^3(\theta)} +C \qquad&\color{comment}\textrm{substituting back } u=\cos(\theta) \\
&=125 \of{-\frac{\sqrt{25-y^2}}{5}+\frac{1}{3}\frac{\of{\sqrt{25-y^2}}^3}{125}} +C \qquad &\color{comment}\textrm{substituting back } \cos(\theta)=\frac{1}{5}\sqrt{25-y^2} \\
&=-25 \sqrt{25-y^2}+\frac{1}{3}\of{\sqrt{25-y^2}}^3 +C \\
\end{split}
\end{equation*}
$$
The following figure summarizes the trigonometric substitution to use with integrals involving $\sqrt{r^2-y^2}$.