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I would use the fact that ln is the inverse function of the exponential function e^x to re-write the equation as x=e^y. This can be directly seen by just putting e^y=e^(lnx). Since the definition of a ln(...

First, use the identity cos(2x)=(cos(x))^2-(sin(x))^2 along with the identity (sin(x))^2+(cos(x))^2=1 to obtain the integral of 1/2*(1-cos(2x)) as it is not possible to integrate (sin(x))^2 straight off w...

Rearrange the second equation in terms of y: meaning that the equation is of the form y = ....-this will give y = 3 - x/2You may now substitute the y in the left hand equation with what y in the right han...

First we must expand the demoninator to; (x+3)(x-3)Then we can multiply the left hand fraction on top and bottom by (x-3) to get a common demoninatorthis gives us; (4x)/((x+3)(x-3)) - ((2)(x-3))/((x+3)(x-...

The important point in the question is the term 'stationary point'. This is where the graph of y will 'flattern out'. If we look at this graph, we can say that the gradient is equal to 0 at this point. Th...