VITEEE 2015 Paper

Let I =

π∕2

∫

0

(log tan x) . sin 2 x dx ... (i)
I =

π∕2

∫

0

log tan (

π

2

−x) sin 2 (

π

2

−x) dx
[Since

a

∫

0

f (x) dx =

a

∫

0

f (a - x) dx]
⇒ I =

π∕2

∫

0

log cot x. sin 2x dx ..(ii)
[Since sin (π - 2x) = sin 2x]
On adding eqs (i) and (ii), we get
2I

π∕2

∫

0

log tan x sin 2 x dx +

π∕2

∫

0

log cot x sin 2x dx
=

π∕2

∫

0

sin 2x log (tan x . cot x) dx
[Since log m + log n = log (m . n)]
=

π∕2

∫

0

sin 2x log dx
⇒ IO = 0 [Since log 1 = 0]
∴

π∕2

∫

0

sin 2x log (tanx) dx = 0