Question

Integration of (secx - 1)^1/2

Answer 1

int. (secx-1)1/2 int. [(1-cosx)/cosx]1/2 Multiply and divide by (1+cosx) Your numerator becomes [(1-cosx)(1+cosx)]1/2 = (1- cos2x)1/2 = (sin2x)1/2 = sinx and your denominator is [(1+cosx)/cosx]1/2 Put cosx = t. ..... -sinxdx = dt Putting these values in the equation, you'll get int. -1/ t(1+t) let -1/t(1+t) = A/t + B/t+1 A = -1, B = 1 So, this becomes .. int. (-1/t) + (1/1+t) Int. -1/t + Int. (1/1+t) -logt + log(1+t) + c log [(1+t)/t] + c Now, put the value of t = cosx Your final ans would be.. log [(1+cosx)/cosx] = log (secx+1) + c