Integration of $\sec^4 x$ While practicing for the AP exam, I came across an integral that I found interesting and I attempted to do by hand: $$\int(\sec^4 x)\, dx$$
Eventually I got stuck, but here are the steps I took-
$$\int(\sec^4 x) \,dx$$
$$\int(\sec^2 x)(\sec x)(\sec x)\,dx$$ 
$$\int(\sec^2 x)(\sec x) (\frac{\tan x}{\sin x})\,dx$$
$$\int(\csc x)(\sec^2 x)(\sec x \tan x)\,dx $$
Apply Integration by parts
$$(u=\csc x, dv=\sec^2 x(\sec x\tan x)$$
$$\frac{\sec^3 x}{3\sin x} + \int(\frac{\sec^3 x}{3})(\csc x \cot x) \,dx$$
$$\frac{\sec^3 x}{3\sin x} + {1\over 3}\int(\sec^2 x)(\csc^2 x) \,dx$$
Here is where I get stuck.. I appreciate any help you can offer on where to go from here!
 A: $$\frac{d}{dx}\tan x=\sec^2x.$$
Therefore
$$\frac{d}{dx}\tan^3 x=3\sec^2x\tan^2x=3\sec^4x-3\sec^2x.$$
So
$$\frac{d}{dx}(\tan^3 x+3\tan x)=3\sec^4x.$$
A: $I=\displaystyle\int\sec^4(x)\,dx$
$I =\displaystyle\int \sec^2(x)(1+\tan^2(x))\,dx$
$u=\tan(x)\implies \sec^2(x)\,dx =\,du$
$I = \displaystyle\int1+u^2\,du$
$I = u+\frac{u^3}3+C$
$I = \tan(x)+\frac{\tan^3(x)}3+C$
A: As noted here, we can integrate powers of secant with integration by parts, viz. $$\int\sec^{n+2}x\,dx=\sec^n x\tan x-n\int\sec^n x\tan^2 x \, dx = \frac{\sec^n x \tan x+n \int\sec^n x \, dx}{n+1}.$$This recursion allows us to go from $\int\sec^2 x \,dx=\tan x + C$ to $$\int\sec^4 x \,dx = \frac{\sec^2 x \tan x + 2\tan x}{3}+C.$$
A: Your approach is a little long.
How about $\sec^4x=(\sec^2x)(\sec^2x)=\sec^2x(\tan^2x+1)=\sec^2x\tan^2x+\sec^2x$. Now you can integrate?
In general: Whenever you have an EVEN power of secant, the idea is to "peel off" one $\sec^2x$ and convert all other $\sec^2x$ terms into $tan^2x+1$ terms. It's a little annoying to work out the parenthesis on $(\tan^2x+1)^n$ but that's just algebra. What is going to happen is that you will end up with tangent terms that all are accompanied with a $\sec^2x$ terms, so they are ready to be integrated with the power rule. 
A: For this problem in particular, I would automatically do what @TheIntegrator did, but in general we can find 
$$I_n=\int\sec^nx\ dx$$
$$I_n=\int\sec^{n-2}x\sec^2x\ dx$$
$$I_n=\int(\tan^2x+1)^{n-2}\sec^2x\ dx$$
Applying the substitution $u=\tan x$ gives
$$I_n=\int(u^2+1)^{n-2}du$$
Assuming that $n\geq2$ is an even integer, we can expand our integrand into a series using the binomial formula:
$$(a+b)^m=\sum_{k=0}^m{m\choose k}a^{m-k}b^k$$
Plugging in $a=u^2$, $b=1$, and $m=n-2$ gives
$$I_n=\int\sum_{k=0}^{n-2}{n-2\choose k}(u^2)^{n-k-2}du$$
$$I_n=\int\sum_{k=0}^{n-2}{n-2\choose k}u^{2n-2k-4}du$$
Cleverly interchanging the $\sum$ and $\int$ signs,
$$I_n=\sum_{k=0}^{n-2}{n-2\choose k}\int u^{2n-2k-4}du$$
Integrating $u^{2n-2k-4}$ and plugging in gives 
$$I_n=\sum_{k=0}^{n-2}{n-2\choose k}\frac{u^{2n-2k-3}}{2n-2k-3}$$
$$I_n=\sum_{k=0}^{n-2}{n-2\choose k}\frac{(\tan x)^{2n-2k-3}}{2n-2k-3}+C$$
Keep in mind that this formula only works for positive even values of $n$. If you want to know a general formula that works for all $n$, I can show you, but it's a bit more involved. 
A: In fact, this is quasi immediate.
$$\int\sec^4x\,dx=\int(\tan^2x+1)\,d\tan x=\frac13\tan^3x+\tan x.$$
