∫ − 2 0 1 6 π 2 0 1 5 π ( cos ( x ) + ∣ cos ( x ) ∣ + cos ( ∣ x ∣ ) ) d x = ?
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I = ∫ − 2 0 1 6 π 2 0 1 5 π cos x + ∣ cos x ∣ + cos ∣ x ∣ d x = I 1 + I 2 + I 3 respectively.
LetWe note that c o s x is an even function with a period or cycle of 2 π and that I c = ∫ a b cos x d x = 0 , when b − a = n π , where n is an integer. Then we have:
I 1 = ∫ − 2 0 1 6 π 2 0 1 5 π cos x d x = 0 Since b − a = 4 0 3 1 π
I 2 = ∫ − 2 0 1 6 π 2 0 1 5 π ∣ cos x ∣ d x = ( 2 0 1 5 − ( − 2 0 1 6 ) ) × 2 ∫ 0 2 π cos x d x = 8 0 6 2 sin x ∣ ∣ ∣ ∣ 0 2 π = 8 0 6 2 All half-cycles are positive.
I 1 = ∫ − 2 0 1 6 π 2 0 1 5 π cos ∣ x ∣ d x = 0 = ∫ − 2 0 1 6 π 2 0 1 5 π cos x d x = 0 Note that cos ∣ x ∣ = cos x as both are even Since b − a = 4 0 3 1 π
Therefore, I = I 1 + I 2 + I 3 = 0 + 8 0 6 2 + 0 = 8 0 6 2 .
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∫ − 2 0 1 6 π 2 0 1 5 π ( cos ( x ) + ∣ cos ( x ) ∣ + cos ( ∣ x ∣ ) ) d x = ∫ − 2 0 1 6 π 2 0 1 5 π cos ( x ) d x + ∫ − 2 0 1 6 π 2 0 1 5 π ∣ cos ( x ) ∣ d x + ∫ − 2 0 1 6 π 2 0 1 5 π cos ( ∣ x ∣ ) d x = [ sin ( x ) ] − 2 0 1 6 π 2 0 1 5 π + ∫ − 2 0 1 5 . 5 π 2 0 1 5 . 5 π ∣ cos ( x ) ∣ d x + ∫ 0 2 0 1 5 π cos ( x ) d x + ∫ 0 2 0 1 6 π cos ( x ) d x = [ sin ( x ) ] 0 2 0 1 5 π + [ sin ( x ) ] 0 2 0 1 6 π + 8 0 6 2 ∫ 0 2 1 π cos ( x ) d x = 8 0 6 2