Question

Solve the following question ​

Answer 1

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 ...

  $\displaystyle\int_0^{\sqrt{3}/2}{\dfrac{\cos{y}}{\sqrt{1-\sin^2{y}}}}\,dy=\int_0^{\sqrt{3}/2}{dy}=\left.\arcsin{x}\right|\limits_0^{\sqrt{3}/2}\\\\=\arcsin{(\sqrt{3}/2)}=\boxed{\dfrac{\pi}{3}}$