Logarithm! (6)

Calculus Level 3

0 π / 2 ln ( tan x ) sin 2 x d x = ? \large \int_0^{{\pi}/{2}} \ln(\tan x) \sin 2x \, dx = \, ?


The answer is 0.

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1 solution

I = 0 π / 2 sin 2 x . l o g ( tan x ) . d x = 0 π / 2 sin ( 2 x ) log ( cot x ) . d x By Adding, We get 2 I = 0 π / 2 sin ( 2 x ) ( log ( tan x ) + log ( cot x ) ) . d x 2 I = 0 π / 2 sin ( 2 x ) log ( tan x . cot x ) . d x = 0 π / 2 sin ( 2 x ) . ( log 1 ) . d x [ log 1 = 0 ] I = 0 \large \displaystyle I = \int_0^{\pi/2} \sin 2x . log(\tan x) . dx = \int_0^{\pi/2} \sin (2x) \log (\cot x) .dx\\ \large \displaystyle \text{By Adding, We get }\\ \large \displaystyle 2I = \int_0^{\pi/2} \sin (2x) (\log (\tan x) + \log (\cot x)).dx\\ \large \displaystyle 2I = \int_0^{\pi/2} \sin (2x) \log (\tan x . \cot x).dx = \int_0^{\pi/2} \sin (2x) . (\log 1) .dx\\ \large \displaystyle \left[ \because \log 1 = 0 \right]\\ \large \displaystyle I = \color{#D61F06}{\boxed0}

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