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How do you integrate #int 1/theta^2cos(1/theta)#?

Answer 1

Luke Alvarez

#-sin(1/theta) + C#

Let #u = 1/theta#. Then #du = -1/theta^2 d theta# and #d theta = -du(theta)^2#.

#=int(1/theta^2 cosu *-du (theta)^2) du#

#= -int(cosu)du#

#=-sinu + C#

Resubstitute:

#=-sin(1/theta) + C#

Hopefully this helps!

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Answer 2

Paisley Johnson

To integrate ( \frac{1}{\theta^2}\cos\left(\frac{1}{\theta}\right) ), you can use the substitution method. Let ( u = \frac{1}{\theta} ), then ( du = -\frac{1}{\theta^2}d\theta ). The integral becomes ( -\int \cos(u) du ). Integrating ( \cos(u) ) gives ( -\sin(u) + C ). Substituting back for ( u ) gives ( -\sin\left(\frac{1}{\theta}\right) + C ), where ( C ) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.