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Question Number 203747 by patrice last updated on 27/Jan/24

	Answered by esmaeil last updated on 27/Jan/24

	$I = \int_{0}^{\frac{π}{2}} \frac{x}{1 + c o s x} d x + \int_{0}^{\frac{π}{2}} \frac{s i n x}{1 + c o s x} d x$ $x = u \to d x = d u$ $\frac{d x}{1 + c o s x} = d v \to v = t a n \frac{x}{2}$ $\to I = x t a n \frac{x}{2} - \int_{0}^{\frac{π}{2}} t a n \frac{x}{2} d x +$ $\int_{0}^{\frac{π}{2}} \frac{s i n x}{1 + c o s x} d x =$ ${x t a n \frac{x}{2} + 2 l n (c o s \frac{x}{2}) + l n (1 + c o s x)]}_{0}^{\frac{π}{2}} =$ $\frac{π}{2} + 2 l n (\frac{\sqrt{2}}{2}) - l n 2 = \frac{π}{2} + l n \frac{1}{2} - l n 2$ $= \frac{π}{2} - 2 l n 2$
	Answered by MathematicalUser2357 last updated on 28/Jan/24

	$I = \frac{π}{2} - 2 \ln 2$
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