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and $I = \int_0^\pi {\frac{{\pi - x}}{{1 + \sin (\pi - x)}}} dx = \int_0^\pi {\frac{{\pi - x}}{{1 + \sin x}}} dx$ ………(ii) On adding Equations (i) and (ii), we get$2I = \pi \int_0^\pi {\frac{1}{{1 + \sin x}}} dx$ $ = \pi \int_0^\pi {\frac{{(1 - \sin x)dx}}{{(1 + \sin x)(1 - \sin x)}}}…
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