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Q.

∫0π/4ln(1+tanx)  dx=

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a

π8log2

b

π4log2

c

πlog2

d

π2log2

answer is A.

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Detailed Solution

I=∫0π/4log(1+Tan x)dx     =∫0π/4log(1+Tan(π4−x))dx    =∫0π/4log(21+Tan x)dx =∫0π/4log(2)-∫0π/4log(1+tanx)dx    =π4log2−I2I=π4log2  I=π8log2
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∫0π/4ln(1+tanx)  dx=