Using integration by parts, and given f(x) = 3xcos(x), find integrate(f(x) dx) between (pi/2) and 0.

We begin by quoting the integration by parts formula, as the question speciaficaly asks us to use it.

integrate(u(x) v'(x) dx)|^(b)(a) = [u(x) v(x)]^(b)(a) - integrate(u'(x) v(x) dx)|^(b)_(a)

To use this formula, we therefore have to split f(x) = 3xcos(x) into the product of u(x) and v'(x), with a view of differentiating u(x), and integrating v'(x). The natural split for f(x) is into 3x and cos(x). Now, for integrating by parts questions, if you are able to 'get rid of a power of x into a constant', i.e. produce a constant by differentiating u(x), then it is a good idea to do it. So in this case we see that selecting u(x) = 3x will allow us to get the constant 3 when we differentiate. We also check that v'(x) = cos(x) can be integrated easily, which is can. Therefore we select

u(x) = 3x, v'(x) = cos(x)

and by differentiating u and integrating v we see

u'(x) = 3, v(x) = sin(x)

The question also gives us that we are integrating between (pi)/2 and 0. Therefore the limits of our integration will be

a = 0, b = (pi)/2

We now substitute these expressions into our by parts formula, so

integrate(u(x) v'(x) dx)|^(b)(a) = [u(x) v(x)]^(b)(a) - integrate(u'(x) v(x) dx)|^(b)_(a)

becomes

integrate(3xcos(x)dx)|^(pi/2)(0) = [3xsin(x)]^(pi/2)(0) - integrate(3sinx dx)|^(pi/2)_(0)

The LHS is what we want to find. Considering the RHS, we yeild

3(pi/2)sin(pi/2) - 3(0)sin(0) - [-3cos(x)]^(pi/2)_(0) = ((3pi)/2)sin(pi/2) - (-3cos(pi/2) - - 3cos(0))

as the integral of sinx is -cosx. Now, sin(pi/2) = 1, sin(0) = 0, cos(pi/2) = 0, cos(0) = 1 are simply trig identities. Therefore we are left with

 (3pi)/2 - (0 + 3) = 3( (pi/2) - 1)

as our final answer, as this is equal to 

integrate(3xcos(x)dx)|^(pi/2)(0)  = integrate(f(x) dx)|^(pi/2)(0)

as required.