How do we integrate [ e^x. cos(x/2 + ?/4 ) between 0 and 2? ?

Asked By aashka On 03/15/2017 07:29

Answers

We'll use a couple of rounds of integration by parts. Let u = e^x, dv = cos(x/2 + pi/4), so du = e^x, v = 2sin(x/2 + pi/4). Our new expression is 2*e^x*sin(x/2+pi/4) - 2*integral(e^x*sin(x/2+pi/4). Now it isn't clear we've gotten anywhere, but let's go one more time on that integral we created. Let u = e^x dv = 2*sin(x/2 + pi/4), so du = e^x and v = -4*cos(x/2 + pi/4). Our new expression is -4*e^x*cos(x/2+pi/4) - (-4)*integral(cos(x/2+pi/4)*e^x). So now here's where we've gotten. integral(e^x*cos(x/2+pi/4)) = 2*e^x*sin(x/2+pi/4) - (-4)*e^x*cos(x/2 + pi/4) - (-4)*integral(e^x*cos(x/2+pi/4)). Notice that the integral is the same on both sides, so let's solve for it. 5*integral(e^x*cos(x/2+pi/4)) = 2e^x*sin(x/2+pi/4) + 4*e^x*cos(x/2+pi/4). So = 1/5*e^x*(2sin(x/2+pi/4) + 4cos(c/2+pi/4)). This can easily be evaluated between whatever limits of integration you like.
Answered On 03/15/2017 13:00