The #dx# at the end of an integral means that the integral should be performed with respect to #x#. So for an equation like #bcos(theta)# which does not have an #x#, the integral is trivial, we just multiply in an #x# and add the constant.
#intbcos(theta)dx = xbcos(theta) + C#
In case the question was actually meant to be #intbcos(theta)d theta# then the answer is still not too hard.
The integral of #cos(x)# is #sin(x)#. Applying this rule, the answer is easily found:
#intbcos(theta)d theta = bintcos(theta) d theta = bsin(theta) + C#