Need help understanding this integration by substitution: $\int \sec^{2}{x} \tan^{3}{x} \, \mathrm{d}x$ The following is from page 142 in chapter 7 of Barron's E-Z Calculus (formerly Calculus the Easy Way), fifth edition:

Evaluate:
43. $y = \int \sec^{2}{x} \tan^{3}{x} \, \mathrm{d}x$

WolframAlpha and the Maxima CAS both agree that the correct answer is $\frac{\tan^{4}{x}}{4} + C$. However, I don't understand how this answer was reached.
I tried the substitution $u = \tan{x}, u' = \sec^{2}{x}$, but that still leaves $\tan^{2}{x}$ in the integral.
What is the correct substitution, and how does it lead to the answer? Where did the 4 in the denominator come from?
 A: We can set $z = \tan x$, and we get that $dz = \sec^2 x dx$. We can then substitute them in to get that $$\int \sec^2 x \tan^3 x dx = \int z^3 dz = \frac{z^4}{4} + C$$ Substitute back in, and we're done.
A: Take $\displaystyle\int{\sec^2{x} \tan^3 {x}} dx$ and note that this is the same as $\displaystyle\int{\tan^3 {x}\sec^2{x} } dx$. 
Knowing that $\displaystyle\frac{d}{dx}{\tan{x}}=\sec^2{x}$, then $\displaystyle\frac{d}{dx}{f(\tan{x})}=f'(\tan{x})\sec^2{x}$ using the chain rule.
Finding the integral of $\sec^2{x} \tan^3 {x}$ is finding such a class of functions whose derivatives are all $\sec^2{x} \tan^3 {x}$. 
Knowing that the power rule is given by $\displaystyle\frac{d}{dx}x^n=nx^{n-1}$, then the power of $\tan x$ in the integral indicates the function from which it came. In this case, $\tan$ takes power 3, so the integral will be power 4. This applies to any composite function.
Using the law of constants (constant multiplier in = constant multiplier out), we can write 
$\displaystyle\int{\tan^3 {x}\sec^2{x} } dx$ = $\displaystyle\frac{1}{4}\displaystyle\int{4\tan^3 {x}\sec^2{x} } dx$
and then seeing that the expression inside the integral is the derivative of an $f(\tan{x})$ we have 
$\displaystyle\frac{1}{4}\displaystyle\int{4\tan^3 {x}\sec^2{x} } dx$ = $\displaystyle\frac{1}{4}(\tan^4 {x} + k)=\displaystyle\frac{1}{4}\tan^4 {x} + C$
recognising in our head that $\displaystyle\frac{d}{dx}(\tan^4{x}+k)=4\tan^{4-1}{x}\sec^2{x}$.
A: Here is a slightly different approach :
Make the following substitution :
$$t=\tan(x) \,\,\,(♣)$$.
$$\implies t^3=\tan^3(x) \,\,\,(♦)$$ 
Differentiate both sides of $(♦)$ to get :
$$3t^2dt =3\tan^2(x) . \sec^2(x)$$
Cancel $3$ from both sides to get :
$$t^2dt=\sec^2(x).\tan^2(x) dx\,\,\,(♠)$$
Now take the integral :
$$y=\int \sec^2(x)\tan^3(x)dx=\int tan(x) \times  \left(\sec^2(x)tan^2(x)\right)dx \,\,\,(♥)$$
Using $(♣)$ and $(♠) , \,\,\,(♥)$ becomes :
$$\int (t) \times (t^2 dt) = \int t^3 dt =\dfrac{t^4}{4}+C\,\,\,\,(♫)$$
Plugging $(♣)$ into $(♫)$, we get :
$$y=\dfrac{\tan^4(x)}{4} + C$$
Hope this helps. $:)$
NOTE : My approach is slightly different from @2012ssohn . He differentiates $u=tan(x)$ whereas I differentiate $\tan^3(x)$. It makes almost no difference. It is just a slightly different way of looking at the same problem.
