$ \int_{0}^{2 n \pi}\left\{|\sin x|-\left|\frac{1}{2} \sin x\right|\right\} d x $ equals:

Question

$ \int_{0}^{2 n \pi}\left\{|\sin x|-\left|\frac{1}{2} \sin x\right|\right\} d x $ equals:, $ n $, $ 2 n $, $ -2 n $, None of these

MARKS App · Accepted Answer

Let I = ∫ 0 2 n π sin x - 1 2 sin x d x = 1 2 ∫ 0 2 n π sin x d x = 1 2 n × ∫ 0 2 π sin x d x Now, I 1 = ∫ 0 2 π sin x d x I 1 = ∫ 0 π sin x d x - ∫ π 2 π sin x d x = - cos x 0 π + cos x π 2 π = - - 1 - 1 + 1 + 1 = 2 + 2 = 4 ∴ I = 1 2 4 n = 2 n

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