Using u-substitution in $\int \tan^3(x) \sec(x)\mathrm{d}x$ $$\int \tan^3(x) \sec(x)\ \mathrm{d}x$$
using the identity $\tan^2(x)=\sec^2(x)-1$
$$\int(\sec^2(x)-1)\sec(x)\tan(x)\ \mathrm{d}x$$
This is where I am confused and to be truthful, I get frustrated with mathematics and truly understand it is not natural and is man made. Why are you allowed to use u-substitution, and it only applies to one value you are substituting?
$$u = \sec(x)$$
$$\mathrm{d}u = \sec(x)\tan(x) \mathrm{d}x$$
 A: $$\int \tan^3(x) \sec(x)\ dx$$
$$=\int (\sec^2(x)-1)\tan(x) \sec(x)\ dx$$
Using $u$-substitution, we have
$$u=\sec(x)$$
$$du=\tan(x)\sec(x)\ dx$$
So now we have
$$\int (u^2-1)\ du=\int u^2\ du-\int du$$
$$=\frac13u^3-u+C=\frac13\sec^3(x)-\sec(x)+C$$
When choosing an expression for $u$, think of it as finding the shortest path to the solution. Many different paths will indeed get you to the solution, but shortest path is often the most efficient.
A: You are allowed u-substitution because it renders the original integral unchanged. Basically what you do in u-substitution is that you change the variable and make it into a more easier terms. If you have ever been into definite integrals, you would have thoroughly understood what changing variable means. You can compute the integral without substitution too as Harish in his unedited post pointed out (by writing $\sec(x)\tan(x)dx=d\sec(x))$ Why would you take the second $\sec (x)$ , u if it more complicates the integral? 
A: u-substitution can be viewed as the chain rule in reverse e.g.
$$\frac{d}{dx} \sin(5x) = \cos(5x) \frac{d}{dx} (5x) = 5 \cos(5x)$$
In reverse:
$$\int 5 \cos(5x) dx = \int \cos(u) du = \sin(u) + C = \sin(5x) + C$$
In your case, if you want, you can replace u in the du expression:
$$u = \sec(x)$$
$$du = \sec(x) \tan(x) = u \tan(x)$$
What is the problem there? Why is the operation worthless just because du contains u? It's just making the integrand easier to integrate. If you like, we can note that
$$\frac{d}{dx} \sin(5x) = 5 \cos(5x)$$
Hence,
$$\int 5 \cos(5x) dx = \sin(5x) + C$$
But if you could conceive of a function (or series) whose derivative is precisely the integrand, you wouldn't need to integrate anything.
In fact, every du will contain u. e.g.
$$\int \frac{1}{x} \ln(x) dx$$
Let $u = \ln(x)$. Then $du = \frac{1}{x} dx = \frac{1}{x} \frac{\ln(x)}{\ln(x)} dx = \frac{1}{x} \frac{u}{u} dx$.
P.S. Mathematics is not natural. It is formal. 
For fun: 1 2
A: $u$-substitution is nothing else but the chain rule for derivation. I mean, if  
$$
F(x) = \int f(x)\, dx \quad \Longleftrightarrow  \quad F'(x) = f(x)
$$
and you have 
$$
x = u(t)
$$
then, chain rule tells us
$$
(F\circ u)'(t) = F'(u(t)) u'(t) \ .
$$
You integrate both sides and get
$$
\int f(x) \, dx = F(x) = (F\circ u) (t) = \int (F\circ u)'(t)\, dt = \int F'(u(t)) u'(t)\, dt \ .
$$
Then, Leibniz notation for the derivative helps us to remember what to do:
$$
\int f(x)\, dx = \left[ x = u(t) \ , dx/dt = u'(t)\  \Leftrightarrow \ dx = u'(t) dt \right] = \int F'(u(t)) u'(t)\, dt \ .
$$
In your case, $u = \sec (x)$ and $du = \tan (x)\sec(x)dx$, so
$$
\int(\sec^2-1) (\sec(x)\tan(x))\mathrm{d}x = \int (u^2 - 1)du \ .
$$
Which is pretty easy to integrate.
A: My simple answer is: because you're multiplying by one. The last step in my example is to make the substitution $u=\sec x$.
$$\begin{array}{lll}
\int (\sec^2x-1)\sec x\tan x dx&=&\int (\sec^2x-1)\sec x\tan x dx\cdot\displaystyle\frac{\frac{d\sec x}{dx}}{\frac{d\sec x}{dx}}\\
&=&\displaystyle\int \frac{(\sec^2 x -1)\sec x\tan xdx\cdot\displaystyle\frac{d\sec x}{dx}}{\displaystyle\frac{d\sec x}{dx}}\\
&=&\displaystyle\int \frac{(\sec^2 x -1)\sec x\tan x d\sec x}{\sec x\tan x}\\
&=&\displaystyle\int (\sec^2 x -1) d\sec x\\
&=&\displaystyle\int (u^2 -1) du\\
\end{array}$$
