If f x pi 40 1 tanx 1 tanx dx evaluate the f x with respect to x ........

Mathematics (NDA) Integration

. If f(x)= ?^π/4₀ (1-tanx)/(1+tanx) dx ,evaluate the f(x) with respect to x?

-log 2

log 2

- ½ log 2

½ log 2

tan(π/4 -x) =1- tan x/1+tanx

integration ?^π/4₀ (1-tanx)/(1+tanx) dx

=?^π/4₀ tan(π/4 -x)

formula ?₀^a f(x)dx =f(a-x)dx

= ?^π/4₀ tan(π/4 -x)

= ?^π/4₀ tan(π/4 –π/4 +x)

= ?^π/4₀ tan x

=[log sec x]₀^π/4

=log sec π/4 -log sec (0)

=log √2 -0

= log √2

= ½ log 2

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