The correct option (b) log 3√4
Explanation:
we know, tan3x = [(tan2x + tanx) / (1 – tanx ∙ tan x)]
∴ tanx + tan2x + tanx ∙ tan2x ∙ tan x – tan3x = 0
∴ I = (π/9)∫0 (tanx + tan2x + tan3x + tanx ∙ tan2x ∙ tan 3x)dx
= (π/9)∫0 2tan3x ∙dx
= 2(π/9)∫0 tan3xdx
= 2[{– log(cos3x)}/3]0(π/9)
= [(– 2)/3] [log cos(π/3) – log(cos 0)]
= [(– 2)/3] log(1/2)
= log(2)–1[(–2)/3]
= log(2)(2/3)
= log(4)(1/3)
= log 3√4.