How do I integrate cos^-1 (x/2) dx??

I feel like this involves a half angle formula, but I am not totally sure. Please help, thank you!

1 Answer
Apr 16, 2018

#intcos^(-1)(x/2)dx=2lnabs(sec(x/2)+tan(x/2))+C#

Explanation:

.

#intcos^(-1)(x/2)dx=int1/cos(x/2)dx=intsec(x/2)dx#

Let #u=x/2, :. du=1/2dx, :. dx=2du#

#intsec(x/2)dx=2intsecudu=2I#

#I=intsecudu#

Let #w=secu+tanu#

#dw=(secutanu+sec^2u)du=secu(secu+tanu)du#

We can write #I# as:

#I=intsecu*(secu+tanu)/(secu+tanu)du#

#I=int(secu(secu+tanu))/(secu+tanu)du#

Now, we substitute:

#I=int(dw)/w=lnabsw+C=lnabs(secu+tanu)+C#

Now, we can substitute for #I and u#:

#intcos^(-1)(x/2)dx=intsec(x/2)dx=2intsecudu=2I=2lnabs(sec(x/2)+tan(x/2))+C#