Can you please show me the process that determines ∫−11dx−x2+1=π? Unfortunately I only know that the derivative of sin−1(x) is equal to the integrand of that, but I don't actually know what to do with it.

Question

Can you please show me the process that determines ∫−11dx−x2+1=π? Unfortunately I only know that the derivative of sin−1(x) is equal to the integrand of that, but I don&#039;t actually know what to do with it.

Kirsten Davis · Accepted Answer

You are right in saying that indeed, ∫11−x2dx=arcsin⁡(x). Now, you need to use the Fundamental Theorem of Calculus, which says that  ∫abf(x)dx=F(b)−F(a)  where F is a function satisfying   F′(x)=f(x)

Alex Sheppard · Accepted Answer

Because you know that  ddxsin−1x=1−x2+1  you are already most of the way there. This tells you that  ∫−11dx−x2+1=sin−1(1)−sin−1(−1)=π2−(−π2)=π  This is due to the Fundamental Theorem of Calculus, which says that for a continuous function f on [a,b], with antiderivative F, then  ∫abf(x)dx=F(b)−F(a)

karton · Accepted Answer

Denote x=sin⁡(t)Because −1≤x≤1 then −π2≤t≤π2. We have:I=∫−111−x2+1dx=∫−π2π21−(sin⁡(t))2+1dsin⁡(t)=∫−π2π2cos⁡tcos⁡tdt=∫−π2π21dtI=π

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

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