∫0π/2 x+sin⁡x1+cos⁡xdx=

∫0π/2 x+sin⁡x1+cos⁡xdx=∫0π/2 x+2sin⁡(x/2)cos⁡(x/2)2cos2⁡(x/2)dx=12∫0π/2 xsec2⁡x2dx+∫0π/2 tan⁡x2dx =12xtan⁡(x/2)(1/2)0π/2−12∫0π/2 tan⁡(x/2)1/2dx+∫0π/2 tan⁡x2dx=π2.

∫0π/2 x+sin⁡x1+cos⁡xdx=

∫0π/2 x+sin⁡x1+cos⁡xdx=∫0π/2 x+2sin⁡(x/2)cos⁡(x/2)2cos2⁡(x/2)dx=12∫0π/2 xsec2⁡x2dx+∫0π/2 tan⁡x2dx =12xtan⁡(x/2)(1/2)0π/2−12∫0π/2 tan⁡(x/2)1/2dx+∫0π/2 tan⁡x2dx=π2.

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$\int_{0}^{π / 2} \frac{x + \sin x}{1 + \cos x} d x =$

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$π / 2$

$π / 4$

$π / 3$

$π / 6$

answer is C.

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Detailed Solution

$\int_{0}^{π / 2} \frac{x + \sin x}{1 + \cos x} d x = \int_{0}^{π / 2} \frac{x + 2 \sin (x / 2) \cos (x / 2)}{2 \cos^{2} (x / 2)} d x = \frac{1}{2} \int_{0}^{π / 2} x \sec^{2} \frac{x}{2} d x + \int_{0}^{π / 2} \tan \frac{x}{2} d x = \frac{1}{2} {[\frac{x \tan (x / 2)}{(1 / 2)}]}_{0}^{π / 2} - \frac{1}{2} \int_{0}^{π / 2} \frac{\tan (x / 2)}{1 / 2} d x + \int_{0}^{π / 2} \tan (\frac{x}{2}) d x = \frac{π}{2} .$