Question

How do you evaluate `int arccosx/sqrt(1-x^2)` from `[0, 1/sqrt2]`?

Answer 1

`int_0^(1/sqrt2)arccosx/sqrt(1-x^2)dx=(3pi^2)/32`

Explanation:

`I=int_0^(1/sqrt2)arccosx/sqrt(1-x^2)dx`

It's important to know that `d/dxarccosx=(-1)/sqrt(1-x^2)`. So, if we let `u=arccosx` then `du=(-1)/sqrt(1-x^2)dx`.

Also note that the bounds will change:

`x=1/sqrt2" "=>" "u=arccos(1/sqrt2)=pi/4`

`x=0" "=>" "u=arccos(0)=pi/2`

Then:

`I=-int_0^(1/sqrt2)arccosx((-1)/sqrt(1-x^2)dx)`

`I=-int_(pi/2)^(pi/4)u color(white).du`

`I=int_(pi/4)^(pi/2)ucolor(white).du`

`I=1/2u^2|_(pi//4)^(pi//2)`

`I=1/2((pi/2)^2-(pi/4)^2)`

`I=1/2(pi^2/4-pi^2/16)`

`I=1/2((4pi^2-pi^2)/16)`

`I=(3pi^2)/32`