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We can rewrite tanxtan2xtan3x in a way that is much easier to integrate. First, find out how else to write tan3x using the tangent angle addition formula.
tan(A+B)=tanA+tanB1−tanAtanB
So, we see that:
tan3x=tan(x+2x)=tanx+tan2x1−tanxtan2x
Cross-multiplying:
(1−tanxtan2x)⋅tan3x=tanx+tan2x
Distributing:
tan3x−tanxtan2xtan3x=tanx+tan2x
Solving for tanxtan2xtan3x:
tanxtan2xtan3x=tan3x−tan2x−tanx
Thus:
∫(tanxtan2xtan3x)dx=∫(tan3x−tan2x−tanx)dx
Splitting this apart:
=∫tan3xdx−∫tan2xdx−∫tanxdx
Note that ∫tanxdx=ln(|secx|)+C. The first two integrals will require using substitution: let u=3x⇒du=3dx and v=2x⇒dv=2dx.
Hence a final answer of:
=ln(|sec3x|)3−ln(|sec2x|)2−ln(|secx|)+C
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