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We have to integrate the following

$$\int \tan x \tan 2x \tan 3x \,\mathrm dx$$

In this I tried as

First split it into sine and cosine terms

Then used $\sin 2x =2\sin x \cos x$

But after that got stuck

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Since $$ \tan3x=\tan(x+2x)=\dfrac{\tan x+\tan2x}{1-\tan x\tan2x}, $$ we have $$ \tan3x-\tan x\tan2x\tan3x=\tan x+\tan2x, $$ i.e. $$ \tan x\tan 2x\tan3x=\tan 3x-\tan x-\tan 2x. $$ For $a\ne 0$ we have $$ \int\tan(ax)\,dx=-\dfrac{1}{a}\ln|\cos(ax)|+c, $$ and therefore \begin{eqnarray} \int \tan x\tan 2x\tan3x\,dx&=&\int(\tan 3x-\tan x-\tan 2x)\,dx\\ &=&\ln|\cos x|+\dfrac{1}{2}\ln|\cos2x|-\dfrac13\ln|\cos3x|+c \end{eqnarray}

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In addition to the other answers, we can also do like this:

Use $\tan 3x =\frac {3\cos^2 x \sin x-\sin ^3 x}{\cos^3 x-3\cos x\sin^2 x}$ ,$\tan 2x=\frac {2\cos x\sin x}{\cos^2 x-\sin^2 x}$ and substitute $u=\cos x $ to get $$I =\int \tan x\tan 2x\tan 3x dx =\int -\frac {2(4\cos^4 x -5\cos^2 x+1)}{\cos x (2\cos^2 x-1)(4\cos^2 x-3)} \sin x dx =2 \int \frac {4u^4-5u^2+1}{u (2u^2-1)(4u^2-3)} du = -\frac {8}{3} \int \frac{u}{4u^2-3} du +\int \frac {2u}{2u^2-1} du +\int \frac {2}{3u} du $$ Hope you can take it from here.

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  • $\begingroup$ Nice approach.. $\endgroup$ – Renascence_5. Jan 7 '17 at 7:28
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Notice $$\tan3x=\frac{\tan2x+\tan x}{1-\tan2x\tan x}\Rightarrow \tan3x\tan2x\tan x=\tan3x-\tan2x-\tan x$$ then $$\int \tan3x\tan2x\tan x \mathrm{d}x = \int \tan3x - \tan2x - \tan x \ \mathrm{d}x$$ the answer will follow.

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(Useful) Trick time!

Note that $\tan 3x = \frac{(\tan 2x + \tan x)}{(1-\tan x \tan 2x)}$. Cross multiply, and you get that $\tan 3x \tan 2x \tan x = -\tan x - \tan 2x + \tan 3x$. Now, the integrals of $\tan x$ is $\log \sec x$, so you can finish it from here.

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