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?
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.