I have to solve this indefinite integration $$\int \frac{\tan^4 \theta d \theta}{1-\tan^2 \theta}$$ 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!



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 in partial fractions and back to $\theta$.

  • $\begingroup$ @Harish Chandra Rajpoot: Thank you for the correction of the lapsus. $\endgroup$ – Bernard May 22 '15 at 22:40
  • $\begingroup$ But $du = \sec^2 \theta d\theta$... $\endgroup$ – Alex G. May 22 '15 at 22:43
  • $\begingroup$ yes you are right. It needs re-editing! $\endgroup$ – Harish Chandra Rajpoot May 22 '15 at 22:44
  • $\begingroup$ @Harish Chandra Rajpoot: finally there was no error in my initial post. I was suggested a correction that I accepted. Only some details were missing. $\endgroup$ – Bernard May 22 '15 at 22:49
  • $\begingroup$ Yes, you are. Sorry I could not point out earlier, you are absolutely right. I re-edited it $\endgroup$ – Harish Chandra Rajpoot May 22 '15 at 22:51

Here is another way to proceed:


$=\displaystyle-\int\sec^2\theta \;d\theta+\int\frac{\frac{1}{2}(1+\cos 2\theta)}{\cos 2\theta}d\theta= -\tan\theta+\int\left(\frac{1}{2}\sec2\theta+\frac{1}{2}\right)d\theta$



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.

  • $\begingroup$ wouldn't it be $\frac{u^4}{1-u^2}$ instead of power $4$? $\endgroup$ – Meow Mix May 22 '15 at 23:30
  • $\begingroup$ @Foliar, No because you multiplied top and bottom by $1+u^2=1+\tan^{2}\theta=\sec^{2}\theta$ $\endgroup$ – TravisJ May 23 '15 at 0:06

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 $\sec(2\theta)$.

  • $\begingroup$ It took me a few moments to guess how you wanted to rewrite $\sin^2 \theta.$ OP might need a hint about that, too. $\endgroup$ – David K May 22 '15 at 22:18
  • $\begingroup$ Yeah fair point - there's so much you could do. But having said that if there's a cos(2x) in the denominator...I guess when you 'see it' it seems so obvious (especially when the result immediately follows)! $\endgroup$ – user140591 May 23 '15 at 1:36

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