Question

What is the value of the integral ${\int }_0^{\frac{\pi }{4}}ln(1+tanx)dx$

• A. $ln\left(2\right).\frac{\pi }{8}$
  
• B. $ln\left(2\right).\pi$
  
  
• C. $ln\left(4\right).\frac{\pi }{4}$
  
  
• D. $ln\left(2\right).\frac{\pi }{3}$

Answer 1

The correct option is A $ln\left(2\right).\frac{\pi }{8}$


I want to say that this is a very famous question. It’s due to the fact that it emphasises the importance of the following property for its evaluation.
${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x).dx.$
So understand all the properties of Definite Integrals and you can use them as an essential tool-kit to tackle this chapter.
So here, let the integral to be evaluated be `I'.
$\therefore {\int }_0^{\frac{\pi }{4}}ln(1+tanx)dx.$
$={\int }_0^{\frac{\pi }{4}}ln(1+tan(\frac{\pi }{4}-x))dx.$
$={\int }_0^{\frac{\pi }{4}}ln(1+\frac{1-tan\left(x\right)}{1+tan\left(x\right)})dx$
$={\int }_0^{\frac{\pi }{4}}ln\left(\frac{2}{1+tan\left(x\right)}\right)dx$
$={\int }_0^{\frac{\pi }{4}}ln\left(2\right)dx-{\int }_0^{\pi /4}ln(1+tan(x\left)\right)dx$
$=ln\left(2\right)x{|}_0^{\frac{\pi }{4}}-l$
$2l=In\left(2\right).\frac{\pi }{4}$
$I=In\left(2\right).\frac{\pi }{8}$