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We will call the integral I, so I = integral of 1/(3+(x+4)^1/2) dx between 0 and 5. First substitute u=3+(x+4)^1/2 into the equation to get I = integral of 1/u dx between 0 and 5 Next ...

First we complete the square on the equation of the circle to obtain: (x-2)^2 -4 + (y+4)^2 -16 = 5. Re arrange : (x-2)^2  + (y+4)^2 = 25 General equation of a circle: (x-a)^2 + (y-b)^2 = r^2, with centre ...

There are several ways to do this which make it particularly interesting. I would advise trying to avoid doing it by eye, this leaves you open to many clumsy mistakes. My preferred method is using equival...