Question

Solve ${\int }_0^{\pi /4}ln(1+tanx)dx$

Answer 1

$I={\int }_0^{\pi /4}ln(1+tanx)dx$
$I={\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx$
$I={\int }_0^{\pi /4}ln(1+tan(\frac{\pi }{4}-x\left)\right)dx$
$I={\int }_0^{\pi /4}ln(1+\frac{tan\frac{\pi }{4}-tanx}{1+tan\frac{\pi }{4}.tanx})dx$
$I={\int }_0^{\pi /4}ln(1+\frac{1-tanx}{1+tanx})dx$
$I={\int }_0^{\pi /4}ln\left(\frac{2}{1+tanx}\right)dx$
$I={\int }_0^{\pi /4}ln\left(2\right)dx-{\int }_0^{\pi /4}ln(1+tan(x\left)\right)dx$
$I={\int }_0^{\pi /4}ln\left(2\right)dx-I$
$2I={\int }_0^{\pi /4}ln\left(2\right)dx$
$I=\frac{ln\left(2\right)}{2}(x{)}_0^{\pi /4}$
$I=\frac{\pi }{8}ln\left(2\right)$