What is #int (cos(1/t))/(t^2)dt #?

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Answer 1 · 2016-02-13T11:51:55

#-sin(1/t) + C#

We can use a substitution. I would recommend using:
#u =1/t#

From this we get:
#du = -1/t^2du#

If we now put this into the integral then it will reduce tot:

#int-cos(u) du#

Which will integrate to give us:

#-sin(u) +C#

Now reverse the substitution:

#-sin(1/t) + C#