Question

π/2Prove that ∫ log tan x dx=0 0

Answer 1

Answer:

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Answer 2

Answer:

0

Step-by-step explanation:

In this question,

We have to evaluate

I = $\int\limits^\frac{\pi}{2}_0 {log\ tanx} \, dx$      .......(i)

I = $\int\limits^\frac{\pi}{2}_0 {log\ tan(\frac{\pi}{2}\ -\ x)} \, dx$

I = $\int\limits^\frac{\pi}{2}_0 {log\ cotx} \, dx$        .......(ii)

Adding equation (i) and (ii) we get,

I + I = $\int\limits^\frac{\pi}{2}_0 {log\ tanx} \, dx + \int\limits^\frac{\pi}{2}_0 {log\ tanx} \, dx$

2I = $\int\limits^\frac{\pi}{2}_0 {log\ tanx\ +\ log\ cotx} \, dx$

2I = $\int\limits^\frac{\pi}{2}_0 {log\ (tanx.Cotx})\, dx$

2I = $\int\limits^\frac{\pi}{2}_0 {log\ tanx.\frac{1}{tanx}} \, dx$

2I = $\int\limits^\frac{\pi}{2}_0 {log1} \, dx$

Since, log1 = 0

2I = 0

I = 0