Q:

What is $\int \sec^3 x \ dx$?

I have to solve for the integral:
$$\int \sec^3 x \ dx$$
I used integration by parts, letting:
$$u=\sec x$$
$$du = \sec x \tan x \ dx$$
$$dv=\sec^2 x$$
$$v=\tan x$$
Integration by parts formula:
$$\int u \ dv = uv - \int v \ du$$
Using integration by parts:
$$\sec x \tan x - \int \tan x \sec x \tan x \ dx$$
$$\sec x \tan x - \int \tan^2 x \sec x \ dx$$
Using the identity $\tan^2 x = \sec^2 x -1$:
$$\sec x \tan x - \int \sec x \left(\sec^2 x - 1\right) \ dx$$
$$\sec x \tan x - \int \sec^3 x \ dx + \int \sec x \ dx$$
$$\sec x \tan x + \ln|\sec x \tan x| - \int \sec^3 x \ dx$$
What do I do now? If I integrate $\int \sec^3 x \ dx$ again, I will just continue going in an infinite loop and end up with $\int \sec^3 x \ dx$ again. Can someone please point me in the right direction? Thanks

A:

Read the Wikipedia page at http://en.wikipedia.org/wiki/Integral_of_secant_cubed - this should give you your answer.
You could continue with what you've done and use the integration constant $C$ to get
$$\int\sec^3(x)dx=\dfrac{1}{2}\left(C+\sec(x)\tan(x)+\ln|\sec(x)\tan(x)|\right)$$

