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Evaluate following.
∫ 0ln (1 + π ) excos(1-ex)dx
Solution
The limits are little unusual in this case, however that will happen sometimes therefore don't get too excited about it. Following is the substitution.
u = 1 - ex du = -ex dx
x =0 ⇒ u = 1 - e0 = 1 -1 = 0
x = ln (1 + π ) ⇒ u = 1 - eln (1+ π ) = 1 - (1 + π ) = - π
Then the integral is,
∫ 0ln (1 + π ) excos(1-ex)dx = ∫0(-∏) cosudu
=-sin u|0(-∏)
= - sin (-∏)-sin0)=0
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