How do you integrate #int e^(sinpix)cospix# from #[0,pi/2]#? Calculus Introduction to Integration Integrals of Exponential Functions 1 Answer Cesareo R. Dec 31, 2016 #1/pi(e^(sin(pi^2/2))-1)# Explanation: Remembering that #d/(dx)e^(sin(pix))=pi cos(pix)e^(sin(pix))# we have #int cos(pix)e^(sin(pix))dx=1/pi(e^(sin(pi^2/2))-1)# Answer link Related questions How do you evaluate the integral #inte^(4x) dx#? How do you evaluate the integral #inte^(-x) dx#? How do you evaluate the integral #int3^(x) dx#? How do you evaluate the integral #int3e^(x)-5e^(2x) dx#? How do you evaluate the integral #int10^(-x) dx#? What is the integral of #e^(x^3)#? What is the integral of #e^(0.5x)#? What is the integral of #e^(2x)#? What is the integral of #e^(7x)#? What is the integral of #2e^(2x)#? See all questions in Integrals of Exponential Functions Impact of this question 2098 views around the world You can reuse this answer Creative Commons License