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I have been tasked with evaluating the integral $$I=\int\frac{\sin(2x)+\sin(4x)-\sin(6x)}{\cos(2x)+\cos(4x)+\cos(6x)+1}dx$$ After substituting first $u=2x$ and then $v=\cos(u)$ and a messy partial fraction decomposition, I get the answer $$\frac{4 \log(\cos(x)) - 2 \log(1 - 2 \cos(2 x)) + 3 \log(\cos(2 x))}{6} + C$$ But given how similar the numerator is to the derivative of the denominator, I suspect there is a much shorter way of going about this involving the identity $$I=\log(\cos(2x)+\cos(4x)+\cos(6x)+1) + \int\frac{3\sin(2x)+5\sin(4x)+5\sin(6x)}{\cos(2x)+\cos(4x)+\cos(6x)+1}dx$$ or something similar and some carefully chosen trigonometric identities. How do I proceed?

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    Are you allowed to use sum/difference formulas for trig functions? Perhaps stuff cancels and the integral becomes much nicer. – imranfat Aug 03 '19 at 18:52
  • Yes, in fact they should be necessary to obtain the simplification in the answer. I have tried but can't see how to do it though. –  Aug 03 '19 at 18:55
  • How about writing it in terms of complex exponentials and trying to simplify? Maybe there is a clever trick to solving this, but this seems to be one of those cases where you don't really learn anything from the exercise, other than slogging though trig identities. – Matematleta Aug 03 '19 at 19:29
  • I tried that, but that gets messy fast (Wolfram Alpha's answer isn't very helpful either ). As for the pedagogical value of this problem, I think there is definitely a science to using such linear combinations to evaluate integrals (see e.g. here) but you may be right that no similarly quick approach would work here. –  Aug 03 '19 at 19:51

3 Answers3

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First express the integrand in the form $$\displaystyle\frac{4\sin 3x \sin 2x \sin x}{4\cos 3x \cos 2x \cos x}.$$ For example, the denominator requires $\cos 2x+\cos 4x\equiv 2\cos 3x \cos x$, $\cos 6x+1\equiv 2\cos^2 3x$, and $\cos 3x+\cos x\equiv 2\cos 2x\cos x$.

Thus the integrand is $\tan 3x\tan 2x\tan x$.

What helps now is the little-known identity $$\tan 3x\tan 2x\tan x\equiv \tan 3x-\tan 2x-\tan x.$$ This starts with $$\sin 3x\sin 2x\sin x \equiv \sin 3x(\cos 2x\cos x-\cos 3x)$$ $$\equiv \sin3x\cos 2x\cos x-\cos3x(\sin 2x\cos x+\cos 2x\sin x),$$ and so on.

Thus the result is $$I=\ln(\cos x)+\frac12\ln(\cos 2x)-\tfrac13\ln(\cos 3x)+c.$$ The identity $\cos x(1-2\cos 2x)\equiv\cos 3x$ implies the equivalence of this answer and the one given above.

The identity $\tan(a+b)\equiv\displaystyle\frac{\tan a+\tan b}{1-\tan a\tan b}$ yields $$\tan a\tan b\tan(a+b)=\tan(a+b)-\tan a-\tan b,$$ so this provides an easier route to the main identity.

urgrue
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  • I saw your answer right after posting mine. I'll leave it up - they're essentially equivalent but your approach is much prettier. I am accepting this answer. –  Aug 04 '19 at 13:29
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    @urgrue - Neat! Just add: normally $\tan 3x = (\tan 2x + \tan x)/(1-\tan 2x \tan x)$ is used in getting the little-known identity, which is more straightforward. – Quanto Aug 04 '19 at 17:38
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The denominator in the integrand of $$I=\int\frac{\sin(2x)+\sin(4x)-\sin(6x)}{\cos(2x)+\cos(4x)+\cos(6x)+1}dx$$

factors as $$2\cos 2x (2\cos 2x -1)(\cos 2x +1)$$

so $u=\cos 2x$ may be helpful.

Mohammad Riazi-Kermani
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I think I may have found a neater approach. Since $$\int \frac{\sin(2x)+\sin(4x)+\sin(6x)}{\cos(2x)+\cos(4x)+\cos(6x)+1} = -\frac{\log(\cos (3x)}{3} + C,$$ we have $$I=-\frac{\log(\cos (3x)}{3} - \int \frac{2 \sin(6x)}{\cos(2x)+\cos(4x)+\cos(6x)+1}.$$ But $$\frac{2 \sin(6x)}{\cos(2x)+\cos(4x)+\cos(6x)+1} = \frac{\sin(3x)}{\cos(x)\cos(2x)} = \frac{\sin(x)\cos(2x)}{\cos(x)\cos(2x)}+\frac{\sin(2x)\cos(x)}{\cos(x)\cos(2x)} = \tan (2x) + \tan (x)$$ so we have $$I=-\frac{\log(\cos (3x))}{3} + \frac{\log(\cos(2x))}{2} +\log(\cos(x)) + C$$ which yields the original answer after simplifying.