∫_0 ^(π/2) (dx/(1+sin x)) →diverges or converges?


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Question Number 132534 by liberty last updated on 15/Feb/21

$\int_{0}^{\frac{π}{2}} \frac{dx}{1 + \sin x} \to diverges or converges ?$
Commented by MJS_new last updated on 15/Feb/21

$we only want to know if it converges or not$ $0 ⩽ x ⩽ \frac{π}{2} \Rightarrow \frac{1}{2} ⩽ \frac{1}{1 + \sin x} ⩽ 1 \Rightarrow converges$
	Answered by Olaf last updated on 15/Feb/21

	$Ω = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + \sin x}$ $Ω = \int_{0}^{\frac{π}{2}} \frac{d x}{1 + \frac{2 t}{1 + t^{2}}} with t = \tan \frac{x}{2}$ $Ω = \int_{0}^{\frac{π}{2}} \frac{1}{1 + t^{2}} . \frac{2 d t}{1 + t^{2}}$ $Ω = - 2 {[\frac{1}{1 + t}]}_{0}^{\frac{π}{2}} = - 2 {[\frac{1}{1 + \tan \frac{x}{2}}]}_{0}^{\frac{π}{2}}$ $Ω = - 2 [\frac{1}{2} - 1] = 1$
	Answered by EDWIN88 last updated on 15/Feb/21
	$We want to evaluate ∫_( 0) ^( (π/2)) (dx/(1+sin x)) . I=∫_0 ^(π/2) (dx/(1+2sin ((x/2))cos ((x/2)))) Wierrstrass Substitution let tan ((x/2))= ∅ ⇒ { ((x=(π/2)→∅=1)),((x=0→∅=0)) :} dx = 2cos^2 ((x/2))d∅ I=∫_0 ^1 (1/(1+2((∅/(1+∅^2 ))))).((2/(1+∅^2 )))d∅ I= ∫_0 ^1 (2/((1+∅)^2 ))d∅ = −2 [(1/(1+∅)) ]_0 ^1 I=−2 [ (1/2)−1 ]= 1 (converges )$
	$We want to evaluate \int_{0}^{\frac{π}{2}} \frac{d x}{1 + \sin x} .$ $I = \int_{0}^{π / 2} \frac{dx}{1 + 2 \sin (\frac{x}{2}) \cos (\frac{x}{2})}$ $Wierrstrass Substitution$ $let \tan (\frac{x}{2}) = \emptyset \Rightarrow {\begin{cases} x = \frac{π}{2} \to \emptyset = 1 \\ x = 0 \to \emptyset = 0 \end{cases}$ $dx = 2 \cos^{2} (\frac{x}{2}) d \emptyset$ $I = \int_{0}^{1} \frac{1}{1 + 2 (\frac{\emptyset}{1 + \emptyset^{2}})} . (\frac{2}{1 + \emptyset^{2}}) d \emptyset$ $I = \int_{0}^{1} \frac{2}{{(1 + \emptyset)}^{2}} d \emptyset = - 2 {[\frac{1}{1 + \emptyset}]}_{0}^{1}$ $I = - 2 [\frac{1}{2} - 1] = 1 (converges)$
	Commented by liberty last updated on 15/Feb/21

	Answered by Ar Brandon last updated on 15/Feb/21

	$\int_{0}^{\frac{π}{2}} \frac{1 - sinx}{\cos^{2} x} dx = {[tanx - \frac{1}{cosx}]}_{0}^{\frac{π}{2}} = 1$
	Commented by EDWIN88 last updated on 15/Feb/21

	$\frac{1}{\tan \frac{π}{2}} - \frac{1}{\cos \frac{π}{2}} = ? \Rightarrow \frac{1}{\infty} - \frac{1}{0} ?$
	Commented by Ar Brandon last updated on 15/Feb/21
	$lim_(x→(π/2)) {tanx−(1/(cosx))}=lim_(x→(π/2)) {((sinx−1)/(cosx))}=lim_(x→(π/2)) {((cosx)/(−sinx))}=0$
	$\lim_{x \to \frac{π}{2}} {tanx - \frac{1}{cosx}} = \lim_{x \to \frac{π}{2}} {\frac{sinx - 1}{cosx}} = \lim_{x \to \frac{π}{2}} {\frac{cosx}{- sinx}} = 0$
	Commented by Ar Brandon last updated on 15/Feb/21

	$\tan \frac{π}{2} - \frac{1}{\cos \frac{π}{2}} = \frac{1}{\cos \frac{π}{2}} - \frac{1}{\cos \frac{π}{2}} = 0$
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