Question

By using properties of definite integrals, evaluate the integrals
${\int }_0^{\frac{\pi }{4}}log(1+tanx)dx.$

Answer 1

Let $I={\int }_0^{\frac{\pi }{4}}log(1+tanx)dx\Rightarrow I={\int }_0^{\frac{\pi }{4}}log[1+tan(\frac{\pi }{4}-x)]dx[\because {\int }_0^{a}f(x)dx={\int }_0^{a}f(a-x\left)dx\right]=f_0^I\frac{\pi }{4}log(1+\frac{1-tanx}{1+tanx})dx[\because tan(A-B)=\frac{tanA-tanB}{1+tanAtanB},hereA=\frac{\pi }{4},B=x]={\int }_0^{\frac{\pi }{4}}log\left(\frac{2}{1+tanx}\right)dx={\int }_0^{\frac{\pi }{4}}\{log2-log(1+tanx\left)\right\}dx[\because log\left(\frac{m}{n}\right)=logm-logn]$

On adding Eqs. (i) and (ii), we get
$2I={\int }_0^{\frac{\pi }{4}}log2dx=log2{\int }_0^{\frac{\pi }{4}}1dx=log2[x{]}_0^{\frac{\pi }{4}}=(log2)(\frac{\pi }{4}-0)\Rightarrow 2I=\frac{\pi }{4}log2\Rightarrow I=\frac{\pi }{8}log2$