How do you find the integral of #int (cos(1/t))/(t^2)dt#?
1 Answer
Apr 26, 2018
# int \ (cos(1/t))/t^2 \ dt = -sin(1/t) + C #
Explanation:
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 #