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Asked Nov 22, 2019
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6. Evaluate the integral.
5T/2
1 cos 2t
dt
2
0
5π/2
1 + cos 2t
dt =
(Type an exact answer, using t as needed.)
2
0
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6. Evaluate the integral. 5T/2 1 cos 2t dt 2 0 5π/2 1 + cos 2t dt = (Type an exact answer, using t as needed.) 2 0

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Expert Answer

Step 1

Given integral is 

5я/2
5я2
1 cos 2t
(1+ cos 2t )dt
2
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5я/2 5я2 1 cos 2t (1+ cos 2t )dt 2

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Step 2

We know that integral of cos...

5/2
sin 2t
5я/2
sin5
1 [ 5л
sin 0
-0
1cos 2t
1
dt
2
22
2
2
2
2
lo
0
5л/2
1cos 2t
-dt
2
57T
4
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5/2 sin 2t 5я/2 sin5 1 [ 5л sin 0 -0 1cos 2t 1 dt 2 22 2 2 2 2 lo 0 5л/2 1cos 2t -dt 2 57T 4

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Math

Calculus

Integration