How do you find integral of ((secxtanx)/(secx-1))dx?

2 Answers
Tom
Apr 13, 2015

int(sec(x).tan(x))/(sec(x)-1)dx

We know that :

sec(x) = 1/cos(x)

tan(x) = sin(x)/cos(x)

So, we have :

=int1/cos(x).sin(x)/cos(x)*1/(1/cos(x)-1)dx

= int(sin(x))/(cos(x)-cos^2(x))dx

Let's t=cos(x)

So dt = -sin(x)

= int1/(t-t^2)(-dt)

Factorize
- int1/(t(1-t))dx

With decomposition in simple elements, we have :

alpha/t+beta/(1-t)

= (alpha(1-t))/(t(1-t))+(betat)/(t(1-t))

So :

alpha-alphat+betat
alpha+t(-alpha+beta)

We have :

alpha = 1
-alpha+beta=0

alpha=beta=1

We get :

-int1/tdt+int1/(1-t)dt

-[log(t)]+[log(1-t)]

=-[log(cos(x))]+[log(1-cos(x))]

=2log(sin(x/2))-log(cos(x))+C

Antoine
Apr 13, 2015

Another way of doing this is to consider that d(secx) = secxtanx*dx

That is, the derivative of secx is secxtanx

=> int(secxtanx)/(secx - 1)dx = int d(secx)/(secx - 1)

This is the same as letting u = secx

We then have,

int(du)/(u - 1) = ln(u - 1) + C = ln(secx - 1) + C

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