Question

How to calculate this? `int_0^piarcsin(sinx)dx.`

Answer 1

Split the integral.

Explanation:

For `x in [0,pi/2]`, we have `arcsin(sinx) = x`

For `x in [pi/2,pi]`, we have `arcsin(sinx) = pi-x`

Therefore,

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

Evaluate the integrals or look at the graph of `y = arcsin(sinx)` between `0` and `pi`

This is a triangle with base `pi` and height `pi/2`, so the area is `pi^2/4`