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Question

Find the value of I if I=π40In(1+tan x)dx


Solution

I=π40In(1+tan x)dx=π40In[1+tan(π4x)dx[Usinga0f(ax)dx]=π40In[1+tanπ4tanx1+tanπ4tan x]dx=π40In[1+1tan xtan x]dx=π40In[21+tanx]dx=π40[In 2In(1+tan x)]dx=π40In 2dxπ40In(1+tan x)dxI=(In 2)[x]π40I2I=π4In 2I=π8In 2

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