Question

Integrate log (tanx) dx​

Answer 1

Answer:

Step-by-step explanation:

∫ ln(tanx)dx , Assume log without any base to be equal to log base e

This Question is solved by definate Integrals, Because Indefinate is very complicated.

Consider, A =   $\int\limits^{\frac{\pi}{2}}_0 {ln(tan x)} \, dx$

A  =        $\int\limits^{\frac{\pi}{4}}_0 {ln(tan x)} \, dx$   +   $\int\limits^{\frac{\pi}{2}}_{\frac{\pi}{4}} {ln(tan x)} \, dx$

In the second integral, substitute, Substituing x = π/2 -T

=> dx =  -dT and we get new form of A as,

A = $\int\limits^{\frac{\pi}{4}}_0 {ln(tan x)} \, dx$    -    $\int\limits^0_{\frac{\pi}{4}} {ln(tan {\frac{\pi}{2} - T})} \, dT$

A = $\int\limits^{\frac{\pi}{4}}_0 {ln(tan x)} \, dx$    +   $\int\limits^0_{\frac{\pi}{4}} {ln(cot T)} \, dT$

A =  $\int\limits^{\frac{\pi}{4}}_0 {ln(tan x)} \, dx$   -   $\int\limits^{\frac{\pi}{4}}_0 {ln(tan T)} \, dT$

We know that,  changing the name of the variable in definite integral doesn’t affect it’s evaluation result.

Hence above step is same as,

A = $\int\limits^{\frac{\pi}{4}}_0 {ln(tan x)} \, dx$  - $\int\limits^{\frac{\pi}{4}}_0 {ln(tan x)} \, dx$

A = 0

=> A =   $\int\limits^{\frac{\pi}{2}}_0 {ln(tan x)} \, dx$ = 0

Thanks.

Hope this helps.