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

Question

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.

Answer 1

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:

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

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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.

Answer 2

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

$\implies \cos t\ge0$

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