Integral of ∫ (1-sinx)/(cosx) dx?

Question

Did the following: 1/cosx - sinx/cosx dx, giving me: ∫secx - tanx dx, is this the right procedure? The book has a different answer...

Yeah, that's the answer. However, I don't see how the devil they got ln(1-sinx) after ln |secx + tanx| - ln |secx| D:

Darn, is it because it's a rule of logarithms? For example log y - log x = log (y/x)?, Hence: ln |secx + tanx| - ln |secx| = ln (secx/secx + tanx/secx)??? Too sleepy and can't think straight... ._.;

Answer 1

You are right so far

Remember that
∫secx dx = ln(sec x + tanx) + c1
∫tanx dx = ln(sec x) + c2

∫secx - tanx dx
= ln(sec x + tan x) - ln(sec x) + c
= ln((sec x + tan x)/(sec x)) + c
= ln(1 + sin x) + c


------------------

∫secx dx
d/dx(sec x) = secx.tanx
d/dx(tan x) = sec^2(x)
∫(sec^2x+secx.tanx)/(sec+tanx) dx
= ln(secx + tanx)


∫tanx dx
= ∫sinx/cosx dx
u = cos x
-sinxdx = du
∫-(1/u)du
= -ln(u)
= ln(1/u)
= ln(1/cosx)
= ln(secx)

Answer 2

L=S(1-sinx)/cosxdx =S(1/cosx-sinx/cosx)dx =S(cosx/1-sin ^2 x)dx -S (sinx/cosx)dx= J - K
* J:
U=sinx => du=cosxdx
=>j=S du/(1-u2) =S du/(1-u)(1+u)= 1/2S du *(1/1-u + 1/1+u)
=1/2ln I 1+u/1-u I + const = 1/2 ln I 1+sinx/1-sinx I + const
*k
k=S -(cosx')/cosx dx= -ln IcosxI +const
=>L

Answer 3

yes. it is right.
integral (secx - tanx)dx=integral(secx)dx - integral(tanx)dx
=[log(secx+tanx)] -[-log(cosx)] + c
=log(secx + tanx) + log(cosx) + c
=log[(secx + tanx)(cosx)] + c
=log(1+sinx) + c .............c=constant of integration
is this the answer in the book?

Answer 4

∫ (1 - sinx) / (cosx) dx =
∫ (1 / cosx - sinx / cosx) dx =
ln(cos(x)) - ln[cos(x/2) - sin(x/2)] + ln[cos(x/2) + sin(x/2)]