Just applying Integration by Substitution for the given equation (Method#1 & Method#2), Let,

$$F(x)=\int \left ( \frac{dx}{\sqrt[]{a^{2}-x^{2}}} \right )\\\tag{1}$$

$\underline{Method \ No.\ 1:}$

Let, $$ x=acos(\theta)$$




$$\theta= cos^{-1}(\frac{x}{a})\\\tag{2}$$

Putting values of $x$ & $dx$ in Eq.(1),

$$F(x)= \int \left ( \frac{-asin(\theta)d\theta}{\sqrt[]{a^{2}-x^{2}}} \right)$$

$$F(x)=\int \left ( \frac{-asin(\theta)d\theta}{a\sqrt[]{1-cos^{2}\theta}} \right)$$

As, $sin^{2}\theta+cos^{2}\theta = 1$ then,

$$F(x)=(-1) \int \left ( \frac{sin(\theta)}{sin(\theta)} \right )d\theta$$

$$F(x)= (-1) \theta$$

Putting value of $\theta$ from Eq.(2):

$$\boxed{F(x)= - cos^{-1}(\frac{x}{a})}\\\tag{3}$$

$\underline{Method \ No.\ 2:}$

Let, $$ x=asin(\theta)$$




$$\theta= sin^{-1}(\frac{x}{a})\\\tag{4}$$

Putting values of $x$ & $dx$ in Eq.(1),

$$F(x)= \int \left ( \frac{acos(\theta)d\theta}{\sqrt[]{a^{2}-x^{2}}} \right)$$

$$F(x)=\int \left ( \frac{acos(\theta)d\theta}{a\sqrt[]{1-sin^{2}\theta}} \right)$$

As, $sin^{2}\theta+cos^{2}\theta = 1$ then,

$$F(x)= \int \left ( \frac{cos(\theta)}{cos(\theta)} \right )d\theta$$

$$F(x)= \theta$$

Putting value of $\theta$ from Eq.(4):

$$\boxed{F(x)= sin^{-1}(\frac{x}{a})}\\\tag{5}$$

Now which of the Eq.(3) or Eq.(5) yields correct result and why ? As can be seen in the graph below, both have different graphs as functions and not equal.

Wolfram Mathematica gives result $F(x)=tan^{-1}(\frac{x}{\sqrt[]{a^{2}-x^{2}}})$ which by the way matches with Eq.(5) result.

Comparison of Eq.(3) & Eq.(5) graphs

  • 1
    $\begingroup$ $\int f(x)\mathop{dx}=F(x)\color{red}{+c}$ $\endgroup$
    – Bonnaduck
    Nov 29, 2021 at 18:48
  • 1
    $\begingroup$ You are calculating primitives. Both are primitives of the same function as both of them differ a constant ($\pi/2$) $\endgroup$
    – rubikman23
    Nov 29, 2021 at 18:48
  • $\begingroup$ @Bonnaduck Yes but how can we know what is the constant for both equations? (if both are right). $\endgroup$ Nov 29, 2021 at 18:51
  • 2
    $\begingroup$ Those questions make no sense. If you know that $F$ is a primitive of $f$, then (assuming that the domain of $f$ is an interval) the set of all primitives of $f$ is $\{F+c\mid c\in\Bbb R\}$. No constant is better than any other constant. $\endgroup$ Nov 29, 2021 at 18:56
  • 1
    $\begingroup$ Let $\theta=\sin^{-1}(x)$, then $x=sin(\theta)$; now, $\sin(\theta)=\cos(\frac{\pi}{2}-\theta)=x$, so $\cos^{-1}(x)=\frac{\pi}{2}-\theta$; from this, $\sin^{-1}(x)+\cos^{-1}(x)=\theta+\frac{\pi}{2}-\theta=\frac{\pi}{2}$; there, you can particularize for your case and get the constant that relates both expressions. $\endgroup$
    – rubikman23
    Nov 29, 2021 at 18:59

2 Answers 2


In general, if $F(x)$ is an antiderivative of $f(x)$, then

$$\int f(x)\mathop{dx}=F(x)+c,$$

where $c$ is an arbitrary constant. You found:

$$\int \frac 1{\sqrt{a^2-x^2}}\mathop{dx}=\sin^{-1}\left(\frac xa\right)+c_1\quad\text{ and }\quad\int \frac 1{\sqrt{a^2-x^2}}\mathop{dx}=-\cos^{-1}\left(\frac xa\right)+c_2.$$

Indeed, $\sin^{-1}\left(x\right)=-\cos^{-1}\left(x\right)+\frac\pi2$. To see this, let $\theta=\sin^{-1}\left(x\right)$. Then $\sin\theta=x$. By the cofunction identity, we have $\cos(\pi/2-\theta)=x$. Hence, $\frac\pi2-\theta=\cos^{-1}x$. Thus, $\theta=\frac\pi2-\cos^{-1}x$. Since $\theta=\sin^{-1}x$, we have


I did gloss over domain restrictions for $\theta$, but the identity still holds.

  • $\begingroup$ Understood. Appreciate your effort. $\endgroup$ Nov 29, 2021 at 19:01


If $f(x)=g(x)+k, f'(x)=?$

But actually for $$I=\int\dfrac{dx}{\sqrt{a^2-x^2}}$$

set $\arcsin\dfrac xa=t\implies x=a\sin t$ and

$-\dfrac\pi2\le t\le\dfrac\pi2$ using https://en.m.wikipedia.org/wiki/Inverse_trigonometric_functions#Principal_values

$\implies \cos t\ge0$

$$I=\int\dfrac{a\cos t}{|a\cos t|}dt=\int\dfrac{a\cos t}{|a|\cos t}dt=\text{ sign}(a)\int dt=?$$


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