Applying Property 6 (∫abf(x)dx=∫abf(a+b−x)dx) with a=0,b=π shows the integrand reverses sign under x→π−x, which forces the integral to equal its own negative — hence 0.
Step 1. Name the integral. Let I=∫0πecos2xcos3[(2n+1)x]dx.
Step 2. Replace x by π−x (Property 6, a+b−x trick with a=0,b=π).
cos2(π−x)=(−cosx)2=cos2x, so ecos2(π−x)=ecos2x is unchanged.
Step 3. Simplify cos[(2n+1)(π−x)]. Expand: cos[(2n+1)π−(2n+1)x]=cos[(2n+1)π]cos[(2n+1)x]+sin[(2n+1)π]sin[(2n+1)x].
Since 2n+1 is an odd integer, cos[(2n+1)π]=−1 and sin[(2n+1)π]=0, so
cos[(2n+1)(π−x)]=−cos[(2n+1)x].
Cubing: cos3[(2n+1)(π−x)]=−cos3[(2n+1)x].
Step 4. Combine. So the integrand at π−x is ecos2x⋅(−cos3[(2n+1)x]), i.e. exactly the negative of the original integrand.
Step 5. Apply Property 6 and solve for I.
I=∫0πecos2xcos3[(2n+1)x]dx=∫0π(−ecos2xcos3[(2n+1)x])dx=−I.
So 2I=0⇒I=0, for every integer n.
Step 6. Match to the printed options. 0 is option (3).