how do you integrate (x+1)cosx dx, using the symmerty rule and it has limits of (pi/2,-pi/2)

Question

I have been asked: "Use symmerty properties to evaluate efficiantly."

(x+1)cosx dx, with limits pi/2 and - pi/2

Answer 1

(x+1)cosxdx=xcosxdx+cosxdx. xcosxdx can be integrated by parts while cosxdx integrated is sinx.

Let u=x and dv=cosxdx, so du=dx and v=sinx.

d(uv)/dx=vdu/dx+udv/dx; integrating with respect to x: uv=int(sinxdx)+int(xcosxdx).

So int(xcosxdx)=uv-int(sinxdx)=xsinx+cosx.

Applying the limits:

(pi)/2-(-(pi)/2*-1)=(pi)/2-(pi)/2=0. 

xsinx+cosx is symmetrical. If f(x)=xsinx+cosx then f(-x)=(-x)(sin(-x))+cos(-x)=xsinx+cosx, because sin(-x)=-sin(x) and cos(-x)=cosx. Therefore f(x)=f(-x) hence f((pi)/2)=f(-(pi)/2) and f((pi)/2)-f(-(pi)/2)=0.