π

∫

0

x

1+sinx

dx
Using the property

a

∫

0

f (x) dx =

a

∫

0

f (a - x) dx
⇒ I =

π

∫

0

π−x

1+sin(π−x)

dx
=

π

∫

0

π−x

1+sinx

dx ... (2)
Now adding (1) and (2), we get
2I =

π

∫

0

x

1+sinx

dx +

π

∫

0

π−x

1+sinx

dx
=

π

∫

0

π

1+sinx

dx
= π

π

∫

0

1

1+sinx

dx
= π

π

∫

0

(1−sinx)

(1−sin2x)

dx
== π

π

∫

0

(1−sinx)

(cos2x)

dx
= π [

π

∫

0

(

1

cos2x

−

sinx

cos2x

)dx]
= π [

π

∫

0

sec2x−secxtanxdx]
= π [

π

∫

0

sec2xdx−

π

∫

0

secxtanxdx]
= π ([tanx]0π−[secx]0π)
⇒ 2I = π (2)
⇒ I = π
So,

π

∫

0

x

1+sinx

dx = π