Evaluate : 
Let, I =
…(1)
Using the property of definite integral: ![]()
∴ I = ![]()
⇒ I =
{∵ sin (2π – x) = -sin x}
⇒ I = 
⇒ I =
…(2)
Adding equation 1 and 2, we get –
2I = ![]()
Using algebra of integrals we have –
2I = ![]()
⇒ 2I =
= 2π – 0
∴ I = π
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