∫ 0 2 π e cos θ cos ( sin θ ) d θ = ?
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c o s ( s i n x ) = e i s i n x + e − i s i n x / 2
I cant understand second line
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e i s i n x = c o s ( s i n x ) + i s i n ( s i n x ) S o , R e a l p a r t o f e i s i n x = c o s ( s i n x )
Nice method! +1
The real part of the integral integral can be written as I = ∫ 0 2 π e e i θ d θ
Now putting e i θ = z and integrating around the unit circle we get,
I = ∫ ∣ z ∣ = 1 z − i e z d z = 2 π i × ( − i ) = 2 π
Alternate solution could be resorting to Feynman's way of Integration by considering the function F ( x ) = ∫ 0 2 π e x cos θ cos ( x sin θ ) d θ and then proceeding.
Our required integrand is F ( 1 ) .
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Let F ( x ) = e c o s x . c o s ( s i n x ) = R . P . O e c o s x . e i s i n x = R . P . O e c o s x + i s i n x = R . P . O e e i x = R . P . O 1 + e i x + 2 ! e 2 i x + 3 ! e 3 i x + . . . . . . .
R . P . O = = Real Part of
Hence you will get
0 ∫ 2 π F ( x ) = 0 ∫ 2 π 1 + c o s x + 2 ! c o s 2 x + 3 ! c o s 3 x + . . . . . = 2 π + 0 + 0 + . . . . = 2 π