J=∫_0 ^(π/2) ((sin x)/(1+sin x+cos x)) dx


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Question Number 175387 by cortano1 last updated on 29/Aug/22

$J = \underset{0}{\int^{π / 2}} \frac{\sin x}{1 + \sin x + \cos x} d x$
Commented by infinityaction last updated on 29/Aug/22

$J = \int_{0}^{π / 2} \frac{2 \sin \frac{x}{2} \cos \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2} + {2 \cos}^{2} \frac{x}{2}} d x$ $J = \frac{1}{2} \int_{0}^{π / 2} \frac{(\sin \frac{x}{2} + \cos \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}{\sin \frac{x}{2} + \cos \frac{x}{2}} d x$ $J = \frac{1}{2} \int_{0}^{π / 2} d x - \frac{1}{2} \int_{0}^{π / 2} \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\sin \frac{x}{2} + \cos \frac{x}{2}} d x$ $J = \frac{π}{4} - {[\log (\sin \frac{x}{2} + \cos \frac{x}{2})]}_{0}^{π / 2}$ $J = \frac{π}{4} - \log (\sqrt{2})$
	Answered by som(math1967) last updated on 29/Aug/22

	$2 J = \int_{0}^{\frac{π}{2}} d x - \int_{0}^{\frac{π}{2}} \frac{d x}{1 + s i n x + c o s x}$ $2 J = \int_{0}^{\frac{π}{2}} d x - \int_{0}^{\frac{π}{2}} \frac{d x}{1 + 2 s i n \frac{x}{2} c o s \frac{x}{2} + 1 - 2 {s i n}^{2} \frac{x}{2}}$ $2 J = x - \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} \frac{x}{2} d x}{2 {s e c}^{2} \frac{x}{2} + 2 t a n \frac{x}{2} - 2 {t a n}^{2} \frac{x}{2}}$ $2 J = x - \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} \frac{x}{2} d x}{2 (1 + t a n \frac{x}{2})}$ $2 J = {[x + l n (1 + t a n \frac{x}{2})]}_{0}^{\frac{π}{2}}$ $2 J = \frac{π}{2} - l n 2$ $J = \frac{π}{4} - l n \sqrt{2}$
	Commented by cortano1 last updated on 29/Aug/22

	$y e s s i r$
	Answered by Ramanathan last updated on 29/Aug/22

	$\underset{4}{π}$ $\frac{π}{4}$
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