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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!

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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 in partial fractions and back to $\theta$.

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  • $\begingroup$ @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
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Here is another way to proceed:

$\displaystyle\int\frac{\tan^4\theta}{1-\tan^2\theta}d\theta=\int\frac{\tan^4\theta-1}{1-\tan^2\theta}d\theta+\int\frac{1}{1-\tan^2\theta}d\theta=-\int(\tan^2\theta+1)d\theta+\int\frac{\cos^2\theta}{\cos^2\theta-\sin^2\theta}d\theta$

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

$=\displaystyle-\tan\theta+\frac{1}{4}\ln\big|\sec2\theta+\tan2\theta\big|+\frac{1}{2}\theta+C$

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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.

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  • $\begingroup$ 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
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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 $\sec(2\theta)$.

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  • $\begingroup$ 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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