a) t = 3x + pi
dt = 3 dx
x = 0 => t = pi
x = pi/6 => t = pi/2 + pi = 3/2 pi
S_[pi, 3/2 pi] 1/3 cos t dt = 1/3 [sin(3/2 pi) - sin pi ]= (-1 - 0)/3 = -1/3
b) sqrt (x) = t
x = t^2
dx = 2t dt
x = 0 => t = 0, x = pi^2 => t = pi
S_[0,pi] sin t * 2t dt = 2 S_[0,pi] t sin t dt = 2 [ - t cos t - S (- cos t ) ]_[0,pi] =
= 2 [ sin t - t cos t ]_[0,pi] = 2 [ 0 - pi *(-1) ] = 2 pi