The value of the integral ∫0nπ+t (|cos x| + |sin x|) dx is
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a
n
b
2n+sin t+cos t
c
cos t
d
sin t–cos t+4n+1
answer is D.
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Detailed Solution
Since the period of |sin x|+|cos x is π/2∫0nπ+t (|sin x|+|cos x|)dx=2n∫0π/2 (|sin x|+|cos x|)dx+∫0t (|sin x|+|cos x|)dx=2n∫0π/2 (sin x+cos x)dx+∫0t (sin x+cos x)dx=(2n)(2)+sin t−cos t+1=(4n+1)+sin t−cos t.