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Q.

The value of the definite integral, ∫θ1θ2dθ1+tanθ=501πK where θ2=1003π2008 and θ1=π2008 . The value of K equals

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answer is 2008.

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Detailed Solution

∴I=∫θ1θ2dθ1+tanπ2−θ=∫θ1θ2tanθdθ1+tanθ and also I=∫θ1θ2dθ1+tanθ∴2I=∫θ1θ2 dθ=θ2−θ1=1002π2008⇒I=501π2008 Hence, K=2008
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