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Mathematics Question on integral

By using the properties of definite integrals, evaluate the integral: ∫0π4log(1+tanx)dx∫_0^{\frac \pi4} log(1+tanx)dx

Answer

Let II =∫0π4log(1+tanx)dx∫_0^{\frac {\pi}{4}} log(1+tan x)dx ...(1)

∴II= ∫0π4log[1+tan(π4−x)]dx∫_0^{\frac {\pi}{4}} log[1+tan (\frac \pi4-x)]dx [∫0af(x)dx=∫0aƒ(a−x)dx][∫_0^af(x) dx = ∫_0^aƒ(a-x)dx]

⇒II = ∫0π4log[1+tanπ4−tan x1+tanπ4.tan x]dx∫_0^{\frac {\pi}{4}} log[1+\frac {tan\frac \pi4-tan \ x}{1+tan \frac \pi4.tan\ x}]dx

⇒II = ∫0π4log[1+1−tan x1+tan x]dx∫_0^{\frac {\pi}{4}} log[1+\frac {1-tan \ x}{1+tan\ x}]dx

⇒II = ∫0π4log[21+tan x]dx∫_0^{\frac {\pi}{4}} log[\frac {2}{1+tan\ x}]dx

⇒II = ∫0π4log 2 dx−∫0π4log (1+tan x)]dx∫_0^{\frac {\pi}{4}} log\ 2\ dx - ∫_0^{\frac {\pi}{4}} log\ (1+tan\ x)]dx

⇒II = ∫0π4log 2 dx−I∫_0^{\frac {\pi}{4}} log\ 2\ dx-I [From(1)]

⇒2I2I = [x.log 2]0π4[x.log\ 2]_0^{\frac {\pi}{4}}

⇒2I2I = π4log 2\frac {\pi}{4}log\ 2

⇒I = π8log 2\frac {\pi}{8}log\ 2