Calculus: Early Transcendentals, Chapter 7, 7.3, Section 7.3, Problem 30

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).