Integration of tanx tna2x tan3x dx please sir , give the answer this question

∫(tanxtan2xtan3x)dx=ln(|sec(3x)|)3−ln(|sec(2x)|)2−ln(|secx|)+C

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(|sec(3x)|)3−ln(|sec(2x)|)2−ln(|secx|)+C