How do you find the antiderivative of #int cos(pit)cos(sin(pit))dt#?

Answer 1

#1/pisin(sin(pit))+C#

Let's try to simplify the trig function within the trig function by letting #u# be the inside function, that is, #u=sin(pit)#. Differentiating this shows that #du=picos(pit)dt#. (Recall to use the chain rule.)
We currently have #cos(pit)dt# in the integral, so we need the factor of #pi#.
#intcos(pit)cos(sin(pit))dt=1/piintcos(sin(pit))*picos(pit)dt#

Substituting in:

#=1/piintcos(u)du#

This is a common integral:

#=1/pisin(u)+C#
Substituting back in #u=sin(pit)#:
#=1/pisin(sin(pit))+C#
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Answer 2

To find the antiderivative of ∫cos(πt)cos(sin(πt))dt, you can use integration by parts. Let u = cos(sin(πt)) and dv = cos(πt)dt. Then, du = -πsin(πt)sin(sin(πt))dt and v = (1/π)sin(πt). Apply the integration by parts formula:

∫udv = uv - ∫vdu

Substitute the values of u, dv, du, and v into the formula and solve the integral. The antiderivative of the given function is:

(1/π)cos(πt)sin(sin(πt)) + 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.

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.

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.

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.

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