What is the integral of #xcos(x)#?

2 Answers
Dec 20, 2014

You use idea of the integrating by parts:
#int uv'dx = uv - intu'vdx #

#intx cosxdx = #
Let:
#u = x#
#u' = 1#
#v' = cosx#
#v = sinx#

Then:
#intx cosxdx = xsinx - int 1*sinxdx = xsinx - (-cosx) = xsinx+cosx#

Dec 20, 2014

The integral is:
#x*sin(x)+cos(x)+C#
You can get this result Integrating by Parts .
In general if you have the product of two functions #f(x)*g(x)# you can try this method in which you have:
#intf(x)*g(x)dx=F(x)*g(x)-intF(x)*g'(x)dx#

The integral of the product of the two functions is equal to the product of the integral ( #F(x)# ) of the first times the second function ( #g(x)# ) minus the integral of ther product of the integral of the first function ( #F(x)# ) times the derivative of the second function ( #g'(x)# ). Hopefully the last integral should be easier to solve than the starting one!!!

In your case you get (you can choose which one is #f(x)# to help you to make the solution easier ):

#f(x)=cos(x)#

#g(x)=x#

#F(x)=sin(x)#

#g'(x)=1#

And finally:
#intx*cos(x)dx=x*sin(x)-int1*sin(x)dx=x*sin(x)+cos(x)+C#
You can now check your answer by deriving this result.