∫0nπ+w |sin⁡x|dx, where n∈N and 0≤w<π is equal to

∫0nπ+w |sin⁡x|dx, where n∈N and 0≤w

A
(2n + 1) + sin w
B
2n + cos w
C
(2n + 1) - cos w
D
None of these

Solution:

$\begin{array}{l} I = \int_{0}^{nπ + w} | \sin x | dx \\ \begin{aligned} = \int_{0}^{w} | \sin x | dx + \int_{w}^{nπ + w} | \sin x | dx = I_{1} + I_{2} \\ I_{1} & = \int_{0}^{w} | \sin x | dx = \int_{0}^{w} \sin xdx \\ = - [\cos x]_{0}^{w} = - \cos w + 1 = 1 - \cos w \\ I_{2} & = \int_{w}^{nπ + w} | \sin x | dx = n \int_{0}^{π} | (\sin x) | dx \\ = n \int_{0}^{π} \sin xdx = n [- \cos x]_{0}^{π} = 2 n \end{aligned} \end{array}$

so, $I = 1 - \cos w + 2 n = (2 n + 1) - \cos w$

$\int_{0}^{nπ + w} | \sin x | dx, where n \in N and 0 \leq w < π$ is equal to

Solution:

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