Question

Integrate the given question:

$\int \limits _{0}^{ \frac{\pi}{4} } ln(1 + tan \: x) dx$
plzz help me ​

Answer 1

Answer:

Hey mate please refer to the attachment

answer is →π/8 log 2

Answer 2

Answer:

I= $\int\limits^\frac{\pi }{4} _0 {ln(1+tanx)} \, dx$

= $\int\limits^\frac{\pi }{4} _0 {ln(1+ tan(\frac{\pi }{4} - x))} \, dx$

= $\int\limits^\frac{\pi }{4} _0 {ln(1+ \frac{tan\frac{\pi }{4} - tanx }{1 + tan\frac{\pi }{4} tanx} )} \, dx$

= $\int\limits^\frac{\pi }{4} _0{ ln(\frac{\(1+tanx+1-tanx)}{{1+tanx)}} \, dx$

= $\int\limits^\frac{\pi }{4} _0 {ln(\frac{2}{1+tanx} )} \, dx$

= $\int\limits^\frac{\pi }{4} _0 {[ln2 - ln (1+tanx)]} \, dx$

I = ㏑2× $\int\limits^\frac{\pi }{4} _0 {x} \, dx$ - $\int\limits^\frac{\pi }{4} _0 {ln(1+tanx)} \, dx$

I = $\frac{\pi ln2 }{4}$ - I

2I =  $\frac{\pi ln2 }{4}$

I = $\frac{\pi ln2 }{8}$             (Ans.)

Step-by-step explanation: