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+1 vote
1.3k views
in Mathematics by (75.1k points)

∫(tan x + tan2x + tan3x + tanx ∙ tan2x ∙ tan3x)dx for x ∈ [0, π/9] is equal to………… 

(a) (1/3)log2 

(b) log3√4 

(c) 3log2 

(d) 4log√3

1 Answer

+2 votes
by (70.5k points)
selected by
 
Best answer

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.

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