# Integration by Substitution for $\int \left ( \frac{dx}{\sqrt[]{a^{2}-x^{2}}} \right )$ gives two results ? Which is correct and why?

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

$$x^{2}=a^{2}cos^{2}(\theta)$$

$$dx=-asin(\theta)d\theta$$

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

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

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

$$x^{2}=a^{2}sin^{2}(\theta)$$

$$dx=acos(\theta)d\theta$$

$$\frac{x^{2}}{a^{2}}=sin^{2}(\theta)$$

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

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.

• $\int f(x)\mathop{dx}=F(x)\color{red}{+c}$ Nov 29, 2021 at 18:48
• You are calculating primitives. Both are primitives of the same function as both of them differ a constant ($\pi/2$) Nov 29, 2021 at 18:48
• @Bonnaduck Yes but how can we know what is the constant for both equations? (if both are right). Nov 29, 2021 at 18:51
• 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. Nov 29, 2021 at 18:56
• 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. Nov 29, 2021 at 18:59

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

$$\sin^{-1}x=\frac\pi2-\cos^{-1}x.$$

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

• Understood. Appreciate your effort. Nov 29, 2021 at 19:01

Hint

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

$$\implies \cos t\ge0$$

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