Evaluate following.

∫ ₀^{ln (1 + π )}  e^xcos(1-e^x)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 - e^x                du = -e^x dx

x =0               ⇒       u = 1 - e⁰  = 1 -1 = 0

x = ln (1 + π )  ⇒      u = 1 - e^{ln (1+ π )}  = 1 - (1 + π ) = - π

Then the integral is,

∫ ₀^{ln (1 + π )}  e^xcos(1-e^x)dx = ∫₀^(-∏) cosudu

                                  =-sin u|₀^(-∏)

                                  =  - sin (-∏)-sin0)=0