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:

$-sin(1/t) + C$

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