9514 1404 393
Answer:
π/3
Step-by-step explanation:
The given integral does not exist. We assume there is a typo in the upper limit, and that you want the integral whose upper limit is (√3)/2.
It is convenient to make the substitution ...
x = sin(y) . . . . so, y = arcsin(x)
dx = cos(y)·dy
Then the integral is ...
[tex]\displaystyle\int_0^{\sqrt{3}/2}{\dfrac{\cos{y}}{\sqrt{1-\sin^2{y}}}}\,dy=\int_0^{\sqrt{3}/2}{dy}=\left.\arcsin{x}\right|\limits_0^{\sqrt{3}/2}\\\\=\arcsin{(\sqrt{3}/2)}=\boxed{\dfrac{\pi}{3}}[/tex]
