Hi Jacob — this is a standard technique which is easy if you have been taught it, but looks like magic if you haven’t. There is a really nice explanation of the technique at this site: http://www.intmath.com/methods-integration/2-integration-logarithmic-form.php
I only learned integrals in prep for this test. When I came across this, I could not integrate? What technique is used here? Or is it simply problem solving?
How do you do part b when you get 2A+4B=0?
Hi Esther, thanks for your question.
We have written x/((x+4)(x+2)) = A/(x+4) + B/(x+2). That means that A(x+2) + B(x+4) = x. So if we consider the coefficients of x we get Ax + Bx = x, and if we consider the numbers we get 2A + 4B = 0.
I don’t quite understand how you went from A(x+2) + B(x+4) = x to Ax + Bx = x.
Yes it does look a bit funny if you write it like that, doesn’t it? But it’s completely fine.
What we are trying to do is to find the values of two numbers A and B, such that x/((x+4)(x+2)) = A/(x+4) + B/(x+2), leading us the the equation A(x+2) + B(x+4) = x.
Since x is a variable that could range across many values, we need to consider the terms that contain x in them separately from the terms that don’t contain x, and this gives us two equations that A and B have to satisfy: Ax + Bx = x (which means that A + B = 0), and also 2A + 4B = 0. If we solve these two equations for A and B, and plug the answers back into the original equation A(x+2) + B(x+4) = x, it should work out.
why can u ignore cosx for part a
Hi Harry, thanks for your question:
We’re not ignoring cos x. We are saying that if you differentiate (1+sin x) with respect to x, you get cos x. Therefore if you differentiate ln(1+sin x) with respect to x, you get (cos x/(1+sin x)). Therefore the integral with respect to x of (cos x/(1+sin x)) is ln(1+sin x).
thanks very much
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