Integrate $\tan^4(\theta)$ I know the answer but I don't know how to find the proper u-sub for it. I'm told I need to use U-sub for all the integrals. Here is where I end up
$$\int (1-\sec^2(\theta))\,d\theta -\int \tan^2(\theta)\sec^2(\theta)\,d\theta$$
My natural u-sub should be $\tan x$ since $dx$ would be $\sec^2$ but I don't know what to do about the other parts. Can I just u sub one section of the integral?
 A: \begin{align}
& \int \tan^4\theta\,d\theta = \int (\tan^2\theta)(\sec^2\theta-1)\,d\theta \\[10pt]
= {} & \int\tan^2\theta\Big(\sec^2\theta\,d\theta\Big) - \int\tan^2\theta\,d\theta \\[10pt]
= {} & \int\tan^2\theta\Big(\sec^2\theta\,d\theta\Big) - \int(\sec^2\theta-1)\,d\theta \\[10pt]
= {} & \int u^2\,du - \tan\theta + \theta+\text{constant}
\end{align}
A: It is good to know that while working with $$\int\sin^m(x)\cos^n(x)dx$$ where $m,~n$ are integers then if $m+n$ is even number so $t=\tan(x)$ is a good substitution. 
A: Yes. You can use u-substitution on the second integral without using it on the first. 
You'll just need to back-substitute on the second, once integrated.
On the first, note that $\frac{d}{d\theta}(\tan \theta + C) = \sec^2\theta \iff \int \sec^2\theta\,d\theta = \tan \theta + C$. 
$$\int \underbrace{(1-\sec^2(\theta))d\theta}_{\large = \theta - \tan \theta} -\int \tan^2(\theta)\sec^2(\theta)d\theta$$
Using $u = \tan \theta \implies du = \sec^2 \theta$ on the second integral: 
$$= \theta - \tan\theta - \int u^2\,du \tag{$u = \tan\theta$}$$
$$= \theta - \tan \theta - \frac{u^3}3 + C\tag{$u = \tan\theta$}$$
$$ = \theta - \tan \theta - \frac 13 \tan^3 \theta + C$$
A: Here's another substitution option. Substitute $u=\tan{(\theta)}$. Then, $\theta=\arctan{(u)}$, and $\mathrm{d}\theta=\frac{\mathrm{d}u}{1+u^2}$, and:
$$\int\tan^4{(\theta)}\,\mathrm{d}\theta=\int\frac{u^4\,\mathrm{d}u}{1+u^2}.$$
Then apply partial fractions.
