Wednesday, July 27, 2016

calculus - Improper integral: inti0nftyfracsin4xx2dx




I have been trying to determine whether the following improper integral converges or diverges:
0sin4xx2dx



I have parted it into two terms. The first term: 1sin4xx2dx

converges (proven easily using the comparison test), but the second term:
10sin4xx2dx
troubles me a lot. What could I do?


Answer



For the second term, you can re-write your function as sinxxsinxxsin2(x). Note that on [0,1], this function is continuous (by the Fundamental Trig Limit you can extend the function to be defined as f(0)=1 at x=0). But then any continuous function on a closed interval is integrable, so 10sin4(x)x2 converges. Hence the whole integral converges.


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