Evaluate: $\int_{0}^{π} \frac{x}{1 + sinx} dx$

Solution

$Let I = \int_{0}^{π} \frac{x}{1 + sinx} dx . . . . . . . . . (i) Using the property \int_{0}^{a} f (x) dx = \int_{0}^{a} f (a - x) dx \Rightarrow I = \int_{0}^{π} \frac{π - x}{1 + \sin (π - x)} dx = \int_{0}^{π} \frac{π - x}{1 + \sin x} dx . . . . . . . . . (ii)$

Now adding (i) and (ii), we get

$2 I = \int_{0}^{π} \frac{x}{1 + sinx} dx + \int_{0}^{π} \frac{π - x}{1 + sinx} dx = \int_{0}^{π} \frac{π}{1 + sinx} dx = π \int_{0}^{π} \frac{1}{1 + sinx} dx = π \int_{0}^{π} \frac{(1 - sinx)}{(1 - \sin^{2} x)} dx = π \int_{0}^{π} \frac{(1 - sinx)}{(\cos^{2} x)} dx = π [\int_{0}^{π} (\frac{1}{\cos^{2} x} - \frac{sinx}{\cos^{2} x}) dx]$

$= π [\int_{0}^{π} \sec^{2} x - secx tanx dx] = π [\int_{0}^{π} \sec^{2} x dx - \int_{0}^{π} secx tanx dx] = π ({[tanx]}_{0}^{π} - {[secx]}_{0}^{π}) \Rightarrow 2 I = π (2) \Rightarrow I = π So, \int_{0}^{π} \frac{x}{1 + sinx} dx = π$

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Evaluate: $\int_{0}^{π} \frac{x}{1 + sinx} dx$