`int_0^(pi/2) cos(t)/(sqrt(1 + sin^2(t))) dt` Evaluate the integral

You need to perform the following substitution to solve the integral `sin t = u => cos t dt = du => t = arcsin u`

`int_0^(pi/2) (cos t dt)/(sqrt(1 + sin^2 t)) = int_(u_1)^(u_2) (du)/(sqrt(1 + u^2) = ln(u + sqrt(u^2+1))|_(u_1)^(u_2)`

Replacing back u for t yields:

`int_0^(pi/2) (cos t dt)/(sqrt(1 + sin^2 t)) = ln(sin t + sqrt(1 + sin^2 t))|_0^(pi/2)`

`int_0^(pi/2) (cos t dt)/(sqrt(1 + sin^2 t)) = ln(sin (pi/2) + sqrt(1 + sin^2 (pi/2))) - ln(sin (0) + sqrt(1 + sin^2 0))`

` int_0^(pi/2) (cos t dt)/(sqrt(1 + sin^2 t)) = ln(1 + sqrt2) - ln(0 + 1)`

` int_0^(pi/2) (cos t dt)/(sqrt(1 + sin^2 t)) = ln(1 + sqrt2) - ln 1`

` int_0^(pi/2) (cos t dt)/(sqrt(1 + sin^2 t)) = ln(1 + sqrt2) - 0`

` int_0^(pi/2) (cos t dt)/(sqrt(1 + sin^2 t)) = ln(1 + sqrt2)`

Hence, evaluating the definite integral yields  `int_0^(pi/2) (cos t dt)/(sqrt(1 + sin^2 t)) = ln(1 + sqrt2).`