How do you evaluate #int arcsinx/sqrt(1-x^2)# from #[0, 1/sqrt2]#?
2 Answers
Explanation:
Note that
#int_0^(pi/4) u/sqrt(1 - x^2) sqrt(1 - x^2)du#
#int_0^(pi/4) u du#
#[1/2u^2]_0^(pi/4)#
We don't have to reverse the substitution, because we adjusted the bounds of integration. If you evaluated in
#1/2(pi/4)^2#
#pi^2/32#
Hopefully this helps!
This is another, and, in my opinion, simpler method to evaluate the integral. However, the answer below is totally valid, and if it helps, use that method instead of this one.
Explanation:
When it comes to integrating this function, one may be tempted to use
But knowing that
So we can evaluate our integral as