How do you find the integral of #int (cos(1/t))/(t^2)dt#?

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Answer 1 · 2018-04-26T00:45:33

# int \ (cos(1/t))/t^2 \ dt = -sin(1/t) + C #

We seek:

# I = int \ \ (cos(1/t))/t^2 \ dt #

We can perform a substitution, Let:

# u=1/t => (du)/dt = -1/t^2 #

Then we can transform the integral:

# I = int \ \ (cos(u))/(-1) \ du #
# \ \ = -sin u + C #

Then, restoring the substitution, we get:

# I = -sin(1/t) + C #