Integral $\int \frac{\tan^4 \theta \,d \theta}{1-\tan^2 \theta}$? I have to evaluate this indefinite integral $$\int \frac{\tan^4 \theta\, d \theta}{1-\tan^2 \theta}$$
I tried it as follows $$I=\int\frac{(\sec^2 \theta-1)\tan^2 \theta\, d \theta}{1-\tan^2 \theta}=\int\frac{\sec^2 \theta \tan^2 \theta\, d \theta}{1-\tan^2 \theta}-\int\frac{\tan^2 \theta \,d \theta}{1-\tan^2 \theta}$$
First part of integration can be easily solved by substitution but how to solve the second part? Help to solve it by other method if you have. Thanks!
 A: Here is another way to proceed:
$\displaystyle\int\frac{\tan^4\theta}{1-\tan^2\theta}d\theta=\int\frac{\tan^4\theta-1}{1-\tan^2\theta}d\theta+\int\frac{1}{1-\tan^2\theta}d\theta=-\int(\tan^2\theta+1)d\theta+\int\frac{\cos^2\theta}{\cos^2\theta-\sin^2\theta}d\theta$
$=\displaystyle-\int\sec^2\theta \;d\theta+\int\frac{\frac{1}{2}(1+\cos 2\theta)}{\cos 2\theta}d\theta= -\tan\theta+\int\left(\frac{1}{2}\sec2\theta+\frac{1}{2}\right)d\theta$
$=\displaystyle-\tan\theta+\frac{1}{4}\ln\big|\sec2\theta+\tan2\theta\big|+\frac{1}{2}\theta+C$
A: Hint: (This expands the hint of @Bernard... it was what I needed to make it work.)
Note that $\frac{1}{1-\tan^{2}(\theta)}=\frac{1+\tan^{2}(\theta)}{(1-\tan^{2}(\theta))(1+\tan^{2}(\theta))}=\frac{\sec^{2}(\theta)}{1-\tan^{4}(\theta)}$.
So using the suggested substitution $u=\tan(\theta)$ gives you
$$\int \frac{\tan^{4}\theta}{1-\tan^{2}\theta}d\theta= \int \frac{u^{4}}{1-u^4}du=\int -1 + \frac{1}{1-u^4}du.$$
Then, $(1-u^4)=(1-u)(1+u)(1+u^2)$ and can be finished with partial fractions.
A: Hint:
Use the substitution  $u=\tan\theta$, $\;\mathrm d\mkern1.5mu u=(1+u^2)\mkern1.5mu\mathrm d\mkern1mu\theta$. You'll get the integral of the rational function:
$$\int\frac{u^4}{1-u^4}\,=\int\frac{u^4}{(1-u)(1+u)(1+u^2)}\,\mathrm d\mkern1mu u$$
Then, decomposition into partial fractions and back to $\theta$.
A: HINT: Multiply the numerator and denominator by $\cos^2(\theta)$. Rewrite $\cos^2(\theta) - \sin^2(\theta)$ as $\cos(2\theta)$ and $\sin^2(\theta)$ in terms of $\cos(2\theta)$. We know how to integrate $\sec(2\theta)$.
