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3.2(A) · Q64

Q.Integrate: ∫tan⁡3xtan⁡2xtan⁡x dx\int \tan 3x\tan 2x\tan x\,dx

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From tan⁡3x=tan⁡(2x+x)=tan⁡2x+tan⁡x1−tan⁡2xtan⁡x\tan3x=\tan(2x+x)=\dfrac{\tan2x+\tan x}{1-\tan2x\tan x}, cross-multiplying and rearranging gives the identity

tan⁡3x−tan⁡2x−tan⁡x=tan⁡3xtan⁡2xtan⁡x\tan3x-\tan2x-\tan x=\tan3x\tan2x\tan x.

So the integrand equals tan⁡3x−tan⁡2x−tan⁡x\tan3x-\tan2x-\tan x.

∫tan⁡3x dx=−13ln⁡∣cos⁡3x∣\int\tan3x\,dx=-\dfrac13\ln|\cos3x|, ∫tan⁡2x dx=−12ln⁡∣cos⁡2x∣\int\tan2x\,dx=-\dfrac12\ln|\cos2x|, ∫tan⁡x dx=−ln⁡∣cos⁡x∣\int\tan x\,dx=-\ln|\cos x|. …

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