Question

How to integrate ∫ root (secx+1)

Answer 1

Do u mean this int. (secx-1)1/2??

Answer 2

i mean int. (secx+1)^1/2

Answer 3

K
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 = sin x

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