# Can you please show me the process that determines \int_{-1}^1

Can you please show me the process that determines $$\displaystyle{\int_{{-{1}}}^{{1}}}{\frac{{{\left.{d}{x}\right.}}}{{\sqrt{{-{x}^{{2}}+{1}}}}}}=\pi$$? Unfortunately I only know that the derivative of $$\displaystyle{{\sin}^{{-{1}}}{\left({x}\right)}}$$ is equal to the integrand of that, but I don't actually know what to do with it.

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Kirsten Davis
You are right in saying that indeed, $$\displaystyle\int{\frac{{{1}}}{{\sqrt{{{1}-{x}^{{2}}}}}}}{\left.{d}{x}\right.}={\arcsin{{\left({x}\right)}}}$$. Now, you need to use the Fundamental Theorem of Calculus, which says that
$$\displaystyle{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$
where F is a function satisfying
F′(x)=f(x)
###### Not exactly what you’re looking for?
Alex Sheppard
Because you know that
$$\displaystyle{\frac{{{d}}}{{{\left.{d}{x}\right.}}}}{{\sin}^{{-{1}}}{x}}={\frac{{{1}}}{{\sqrt{{-{x}^{{2}}+{1}}}}}}$$
you are already most of the way there. This tells you that
$$\displaystyle{\int_{{-{1}}}^{{1}}}{\frac{{{\left.{d}{x}\right.}}}{{\sqrt{{-{x}^{{2}}+{1}}}}}}={{\sin}^{{-{1}}}{\left({1}\right)}}-{{\sin}^{{-{1}}}{\left(-{1}\right)}}={\frac{{\pi}}{{{2}}}}-{\left(-{\frac{{\pi}}{{{2}}}}\right)}=\pi$$
This is due to the Fundamental Theorem of Calculus, which says that for a continuous function f on [a,b], with antiderivative F, then
$$\displaystyle{\int_{{a}}^{{b}}}{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={F}{\left({b}\right)}-{F}{\left({a}\right)}$$
karton

$$\\\text{Denote}\ x=\sin(t) \\\text{Because}\ -1 \leq x \leq 1\ \text{then}\ -\frac{\pi}{2} \leq t \leq \frac{\pi}{2}.\ \text{We have:} \\I=\int_{-1}^1 \frac{1}{\sqrt{-x^2+1}}dx \\=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{\sqrt{-(\sin(t))^2+1}}d \sin (t) \\=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos t}{\cos t}dt \\=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}1 dt \\I=\pi$$