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Question Number 108667 by bobhans last updated on 18/Aug/20

	Answered by john santu last updated on 18/Aug/20

	Commented by john santu last updated on 18/Aug/20

	$t y p o t h e l a s t l i n e i s \frac{\sqrt{2}}{4} \ln (3 + 2 \sqrt{2})$
	Commented by bobhans last updated on 18/Aug/20

	$y e s . . . . r i g h t$
	Answered by ajfour last updated on 18/Aug/20

	$2 I = \int_{0}^{π / 2} \frac{d x}{\sin x + \cos x}$ $l e t \tan \frac{x}{2} = t \Rightarrow (1 + t^{2}) d x = 2 d t$ $I = \int_{0}^{1} \frac{d t}{2 t + 1 - t^{2}} = \int_{0}^{1} \frac{d t}{{(\sqrt{2})}^{2} - {(t - 1)}^{2}}$ $= \frac{1}{2 \sqrt{2}} \ln ∣ \frac{\sqrt{2} + t - 1}{\sqrt{2} - t + 1} ∣_{0}^{1}$ $= \frac{1}{2 \sqrt{2}} \ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1}) = \frac{1}{\sqrt{2}} \ln (1 + \sqrt{2})$ $I = \int_{0}^{π / 2} \frac{\sin^{2} x}{\sin x + \cos x} d x = \frac{1}{\sqrt{2}} \ln (1 + \sqrt{2}) .$
	Commented by bobhans last updated on 18/Aug/20

	$t y p o . \frac{1}{2 \sqrt{2}} ? = \frac{1}{\sqrt{2}} . a n d \frac{\sqrt{2} + 1}{\sqrt{2} - 1} ? = 1 + \sqrt{2}$
	Commented by ajfour last updated on 18/Aug/20

	$y o u d i n t f o l l o w S i r$ $\frac{1}{2 \sqrt{2}} \ln (\frac{1 + \sqrt{2}}{\sqrt{2} - 1}) = \frac{1}{2 \sqrt{2}} \ln {(1 + \sqrt{2})}^{2}$ $= \frac{1}{\sqrt{2}} \ln (1 + \sqrt{2}) .$
	Answered by mathmax by abdo last updated on 18/Aug/20
	$I =∫_0 ^(π/2) ((sin^2 x)/(sinx +cosx))dx and J =∫_0 ^(π/2) ((cos^2 x)/(sinx +cosx))dx( friend of I) ⇒I−J =∫_0 ^(π/2) ((sin^2 x −cos^2 x)/(sinx +cosx))dx =∫_0 ^(π/2) (sinx −cosx)dx =[−cosx −sinx]_0 ^(π/2) =−1+1 =0 ⇒I =J I +J =2I =∫_0 ^(π/2) (dx/(cosx +sinx)) =_(tan((x/2))=t) ∫_0 ^1 ((2dt)/((1+t^2 )(((1−t^2 )/(1+t^2 ))+((2t)/(1+t^2 ))))) =2 ∫_0 ^1 (dt/(1−t^2 +2t)) =−2 ∫_0 ^1 (dt/(t^2 −2t−1)) Δ^′ =1+1 =2 ⇒t_1 =1+(√2) and t_2 =1−(√2) ⇒ I =− ∫_0 ^1 (dt/((t−t_1 )(t−t_2 ))) =−(1/(2(√2)))∫_0 ^1 ((1/(t−t_1 ))−(1/(t−t_2 )))dt =(1/(2(√2)))[ln∣((t−t_2 )/(t−t_1 ))∣]_0 ^1 =(1/(2(√2)))[ln∣((t−1+(√2))/(t−1−(√2)))∣]_0 ^1 =(1/(2(√2))){−ln∣((−1+(√2))/(−1−(√2)))∣} =−(1/(2(√2)))ln((((√2)−1)/((√2)+1))) =(1/(2(√2)))ln((((√2)+1)/((√2)−1)))$
	$I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{sinx + cosx} dx and J = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{sinx + cosx} dx (friend of I)$ $\Rightarrow I - J = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x - \cos^{2} x}{sinx + cosx} dx = \int_{0}^{\frac{π}{2}} (sinx - cosx) dx$ $= {[- cosx - sinx]}_{0}^{\frac{π}{2}} = - 1 + 1 = 0 \Rightarrow I = J$ $I + J = 2 I = \int_{0}^{\frac{π}{2}} \frac{dx}{cosx + sinx} =_{\tan (\frac{x}{2}) = t} \int_{0}^{1} \frac{2 dt}{(1 + t^{2}) (\frac{1 - t^{2}}{1 + t^{2}} + \frac{2 t}{1 + t^{2}})}$ $= 2 \int_{0}^{1} \frac{dt}{1 - t^{2} + 2 t} = - 2 \int_{0}^{1} \frac{dt}{t^{2} - 2 t - 1}$ $Δ^{^{'}} = 1 + 1 = 2 \Rightarrow t_{1} = 1 + \sqrt{2} and t_{2} = 1 - \sqrt{2} \Rightarrow$ $I = - \int_{0}^{1} \frac{dt}{(t - t_{1}) (t - t_{2})} = - \frac{1}{2 \sqrt{2}} \int_{0}^{1} (\frac{1}{t - t_{1}} - \frac{1}{t - t_{2}}) dt$ $= \frac{1}{2 \sqrt{2}} {[\ln ∣ \frac{t - t_{2}}{t - t_{1}} ∣]}_{0}^{1} = \frac{1}{2 \sqrt{2}} {[\ln ∣ \frac{t - 1 + \sqrt{2}}{t - 1 - \sqrt{2}} ∣]}_{0}^{1}$ $= \frac{1}{2 \sqrt{2}} {- \ln ∣ \frac{- 1 + \sqrt{2}}{- 1 - \sqrt{2}} ∣} = - \frac{1}{2 \sqrt{2}} \ln (\frac{\sqrt{2} - 1}{\sqrt{2} + 1}) = \frac{1}{2 \sqrt{2}} \ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1})$
	Answered by 1549442205PVT last updated on 18/Aug/20

	$Set J = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} x}{sinx + cosx} dx and t = \frac{π}{2} - x$ $\Rightarrow dx = - dt \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{\sin^{2} x}{sinx + cosx} dx$ $= \int_{\frac{π}{2}}^{0} \frac{\cos^{2} t}{sint + cost} (- 1) dt = \int_{0}^{\frac{π}{2}} \frac{\cos^{2} t}{sint + cost} dt = J$ $\Rightarrow 2 I = I + J = \int_{0}^{\frac{π}{2}} \frac{dx}{cosx + sinx}$ $= \int_{0}^{\frac{π}{2}} \frac{cosx + sins}{{(sinx + cosx)}^{2}} dx$ $= \int_{0}^{\frac{π}{2}} \frac{\sqrt{2} \sin (x + \frac{π}{4})}{{[\sqrt{2} \sin (x + \frac{π}{4})]}^{2}} dx$ $= - \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{d [\cos (x + \frac{π}{4})]}{1 - \cos^{2} (x + \frac{π}{4})} dx$ $Set \cos (x + \frac{π}{4}) = u \Rightarrow u ∣_{1 / \sqrt{2}}^{- 1 / \sqrt{2}}$ $= \frac{1}{\sqrt{2}} \int_{0}^{\frac{π}{2}} \frac{du}{u^{2} - 1} dx = \frac{1}{2 \sqrt{2}} \int_{0}^{\frac{π}{2}} (\frac{1}{u - 1} - \frac{1}{1 + u}) du$ $= \frac{1}{2 \sqrt{2}} (\ln ∣ u - 1 ∣ - \ln ∣ u + 1 ∣) ∣_{1 / \sqrt{2}}^{- 1 / \sqrt{2}}$ $== \frac{1}{2 \sqrt{2}} \ln ∣ \frac{u - 1}{u + 1} ∣ ∣_{1 / \sqrt{2}}^{- 1 / \sqrt{2}} = \frac{1}{2 \sqrt{2}} (\ln ∣ \frac{\sqrt{2 +} 1}{\sqrt{2} - 1} ∣ - \ln \frac{\sqrt{2} - 1}{\sqrt{2} + 1})$ $= \frac{1}{2 \sqrt{2}} \ln \frac{{(\sqrt{2} + 1)}^{2}}{{(\sqrt{2} - 1)}^{2}} = \frac{1}{\sqrt{2}} \ln \frac{\sqrt{2} + 1}{\sqrt{2} - 1} = \sqrt{2} \ln (\sqrt{2} + 1)$ $\Rightarrow I = \frac{1}{\sqrt{2}} \ln (\sqrt{2} + 1)$
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