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Problem ∫ 0π(5cost+5) 2 +(−5sint) 2

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First, we define the functions:

Let's start with the function f1 which is 5 cos(t) + 5, and the function f2 which is -5 sin(t).

The next step is to calculate their squares. Thus, we have f1_squared = (5 cos(t) + 5)^2 and f2_squared = (-5 sin(t))^2.

Then, we add these squared functions together.

The sum of these squared functions is (5 cos(t) + 5)^2 + (-5 sin(t))^2.

Now, we are going to compute the integral of this sum with respect to t from 0 to pi.

This gives us ∫ from 0 to π [(5 cos(t) + 5)^2 + (-5 sin(t))^2] dt.

Our computations yield the result that this definite integral is equal to 50π.

So, the integral from 0 to π of the sum of these squared functions, (5 cos(t) + 5)^2 + (-5 sin(t))^2, is equal to 50π.

User Shivanand Darur
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