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Integrate: $$\int \frac{\tan ^3\left(\ln \left(x\right)\right)}{x}dx$$

My attempt:

$$u=\ln(x) \implies \int \tan ^3\left(u\right)du=\int \tan ^2\left(u\right)\tan \left(u\right)du=\int \left(-1+\sec ^2\left(u\right)\right)\tan \left(u\right)du$$

I'm having trouble after this section. My initial idea was to use $u$-subsituition again. So, $$v=\sec(u) \implies \int \frac{-1+v^2}{v}dv=\int \:-\frac{1}{v}+vdv=-\ln \left|v\right|+\frac{v^2}{2}=-\ln \left|\sec \left(\ln \left(x\right)\right)\right|+\frac{\sec ^2\left(\ln \left(x\right)\right)}{2} + c, c \in \mathbb{R}$$

I'm pretty sure this is correct. I'm curious if there is any other way to solve this (hopefully in an easier manner?)?

José Carlos Santos
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Ski Mask
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  • Expand the brackets to obtain $\int - \tan u + sec^{2} u \tan udu$. Then for $\tan u$ is just $-ln \mid \cos u \mid$ while $sec^{2} u \tan u$ is u need to integrate $sec^{2} u$ and change the differential to $\tan u$ so just integrate in terms of $\tan u$. Thus your final answer is $ln \mid \cos u \mid + (\tan u)^{2}/2$. – popping900 Jul 15 '20 at 15:44
  • It looks fine to me. The first change of variable is obvious. Once you get a trigonometric integral involving $\tan$ here may be other quick tricks but your seems natural sn not difficult at all. – Mittens Jul 15 '20 at 15:44
  • Yeah easier method is to solve without substitution – Namburu Karthik Jul 15 '20 at 15:50

2 Answers2

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Note that$$\int\tan^3x\,\mathrm dx=\int\frac{\sin^3x}{\cos^3 x}\,\mathrm dx=\int\frac{\sin(x)\bigl(1-\cos^2(x)\bigr)}{\cos^3x}\,\mathrm dx.$$So, if you do $\cos x=t$ and $-\sin x\,\mathrm dx=\mathrm dt$, you get$$-\int\frac{1-t^2}{t^3}\,\mathrm dt.$$

José Carlos Santos
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$$\int \tan^3 (x) = \int \tan(x) ( \sec^2 (x) -1) dx$$

$$ \implies \int \tan (x) \sec^2 (x) dx -\int \tan (x) dx = \int \tan (x) \mathbb{d}(\tan x) -\int \tan (x) dx $$

Can you finish? :)

This trick generalizes for all odd powers of $\tan (x)$ type integrals

Clemens Bartholdy
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