∫ sin 2 0 1 0 ( x ) sec 2 ( x ) − 2 0 1 0 d x = ( sin ( x ) ) 2 0 1 0 P ( x ) + C
If the equation above is true for some function P ( x ) with P ( 4 π ) = 2 1 0 0 5 1 , where C denotes the arbitrary constant, then find the value of P ( 3 π ) .
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Can you explain what the " " symbol means in the integration?
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Oh actually I wanted to show the terms cancel out
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Ah. Unfortunately you will not be able to do a complete strikethrough using Latex (without having additional packages installed in our backend).
I've edited the problem. Can you edit your solution and indicate why that condition is necessary? IE point out what was incomplete in your solution. If you need help, look at Jake's report.
Just a slight thing to scare people from answering.
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its simple
∫ ( sin x ) 2 0 1 0 sec 2 x d x − ∫ ( sin x ) 2 0 1 0 2 0 1 0 d x
In the former integral apply integration by parts considering ( sin x ) 2 0 1 0 1 as first function and sec 2 x as second
( sin x ) 2 0 1 0 tan x + ∫ ( sin x ) 2 0 1 0 2 0 1 0 d x − ∫ ( sin x ) 2 0 1 0 2 0 1 0 d x
so thus P ( x ) = tan ( x )