How does trigonometrical substituition work for integrating a function?

Question

I am recently learning integration and I came across trigonometrical substituition but I am confused that how is it working? For example: For integrating

$\int \frac{1}{\sqrt{1-x^2}} dx$

We put $x=\sin y$

as per my teacher told, we assume x and a (any constant and here in this case a=1) as sides of a right triangle and y an angle in it and then we substitute the values as per it but I am confused if we can do that then why don't we put

$x=\cos y $

I tried solving the integral substituting $x=\cos y $ but I got the answer negative.

Answer 1

If you do $x=\sin y$ and $\mathrm dx=\cos(y)\,\mathrm dy$, then $\int\frac1{\sqrt{1-x^2}}\,\mathrm dx$ becomes$$\int\frac{\cos(y)}{\sqrt{1-\sin^2(y)}}\,\mathrm dy=\int1\,\mathrm dy=y+C=\arcsin(x)+C.$$And if you do $x=\cos y$ and $\mathrm dx=-\sin(y)\,\mathrm dy$, then $\int\frac1{\sqrt{1-x^2}}\,\mathrm dx$ becomes$$-\int\frac{\sin(y)}{\sqrt{1-\cos^2(y)}}\,\mathrm dy=-\int1\,\mathrm dy=-y+C=-\arccos(x)+C.$$Since $\arcsin+\arccos$ is constant (it is equal to $\frac\pi2$), theses answers are the same.

Answer 2

With $x=\cos y$, you would get the answer as $-\cos^{-1}x+c$, which can be written as $-\cos^{-1}x+\frac{\pi}2+c_1$, where $c=\frac{\pi}2+c_1$. Now, $-\cos^{-1}x+\frac{\pi}2=\sin^{-1}x$, which is the same answer you would have got with $x=\sin y$ substitution.