2007 methods cas exam 1 question 7
f(x) = xcos(3x), it is known that f’(x) = cos(3x)-3xsin(3x)
use this fact to find an antiderivative of xsin(3x)
cos(3x)-3xsin(3x)dx = xcos(3x)
-3xsin(3x)dx = xcos(3x)-(sin(3x))/3
xsin(3x)dx = ((xcos(3x)-(sin(3x))/3))/-3
is my working out and answer correct?
This is very difficult to follow because it's not clear which expressions are integrals, and which aren't. You will be penalised for incorrect notation on the exam.
Is it a general rule that if we’re taking something to the other side, ie. add or subtract, we take the integral along with the term, but if it’s dividing or multiplying, we don’t take the integral, but just the term itself?
I'm not quite sure what you mean. I think what you are getting at with can be answered by referring to the following two properties of integrals:
 + g(x)dx = \int f(x)dx + \int g(x)dx \\<br />\int k f(x)dx = k \int f(x)dx)
Methods exam 1 2017 question 6
Is my working out in part p aligned with "hence"?
Yes, more or less, although this could be solved more elegantly by simply noting that tan(x) is one-to-one on the domain [0, pi], and hence each of the three equations in terms of tan(x) have exactly one solution, so no need to consider the symmetries. There is also no need for the algebraic manipulation of sin^2(x) = 3cos^2(x) to get the two equations for tan(x) – these are already implied by part a.
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