Let I =

π

∫

0

ecosx

ecosx+e−cosx

dx
Using

a

∫

0

f (x) =

a

∫

0

f (a - x) dx
I =

π

∫

0

ecos(π−x)

ecos(π−x)+e−cos(π−x)

dx
2I =

π

∫

0

e−cosx+ecosx

ecosx+e−cosx

dx
I =

1

2

π

∫

0

dx =

1

2

[π - 0] =

π

2

OR
I =

π 2

∫

0

(2 log sin x - log sin 2x) dx
I =

π 2

∫

0

(log‌

sin2x

2sinx.cosx

.dx)
I =

π 2

∫

0

log (

tanx

2

) . dx ... (i)
Using property

a

∫

0

f (x) dx =

a

∫

0

f (a - x) dx
We get,
I =

π 2

∫

0

log (

tan( π 2 −x)

2

) dx
⇒ I =

π 2

∫

0

log (

cotx

2

) dx ... (ii)
Additing (i)&(ii)
2I =

π 2

∫

0

[log(

tanx

2

)+log(

cotx

2

)] dx
⇒ 2I =

π 2

∫

0

log [(

tanx

2

)(

cotx

2

)] dx
⇒ I =

1

2

π 2

∫

0

log (

1

4

) dx
⇒ I =

1

2

log (

1

4

)×(

π

2

)
⇒ I =

1

2

log (

1

4

)

1

2

× (

π

2

)
⇒ I = log (

1

2

)×(

π

2

)
⇒ I =

π

2

‌log‌

1

2