# Explain how is calculate in intervals [0,pi] antiderivative sin^-1(sin x)?

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### 1 Answer

Notice that the principal difficulty of this problem is that the function changes over the interval [0,pi] such that:

`x in [0,pi/2] => arcsin(sin x) = x`

`x in (pi/2,pi] => arcsin(sin x) = arcsin(sin(pi-x)) = pi-x`

Hence, you need to evaluate the definite integral using the property of linearity such that:

`int_0^pi arcsin(sin x)dx = int_0^(pi/2) arcsin(sin x)dx + int_(pi/2) ^pi arcsin(sin (pi-x))dx`

`int_0^pi arcsin(sin x)dx = int_0^(pi/2) x dx +int_(pi/2)^pi (pi-x)dx`

`int_0^pi arcsin(sin x)dx = x^2/2|_0^(pi/2) + pix|_(pi/2)^pi - x^2/2|_(pi/2)^pi`

`int_0^pi arcsin(sin x)dx = pi^2/8 + pi^2 - pi^2/2 - pi^2/2 + pi^2/8`

`int_0^pi arcsin(sin x)dx = 2pi^2/8 - 2pi^2/2 + pi^2`

`int_0^pi arcsin(sin x)dx = pi^2/4 - pi^2 + pi^2`

`int_0^pi arcsin(sin x)dx = pi^2/4`

**Hence, evaluating carefully the definite integral of the function arcsin(sin x) yields `int_0^pi arcsin(sin x)dx = pi^2/4` and not `pi^2/2.`**