∫_0 ^(π/2) (dx/(√(1+sin x))) ?


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Question Number 95848 by john santu last updated on 28/May/20

$\underset{0}{\int^{\frac{π}{2}}} \frac{dx}{\sqrt{1 + \sin x}} ?$
	Answered by bobhans last updated on 28/May/20

	$\underset{0}{\int^{\frac{π}{2}}} \frac{dx}{\sqrt{{(\sin \frac{x}{2} + \cos \frac{x}{2})}^{2}}} = \underset{0}{\int^{\frac{π}{2}}} \frac{dx}{\sqrt{2} \cos (\frac{x}{2} - \frac{π}{4})}$ $set \frac{x}{2} - \frac{π}{4} = t \Rightarrow dx = 2 dt$ $\underset{- \frac{π}{4}}{\int^{0}} \frac{2 dt}{\sqrt{2} \cos t} = \sqrt{2} [\ln ∣ \sec t + \tan t ∣]_{- \frac{π}{4}}^{0}$ $= \sqrt{2} \ln ∣ \frac{1 + \sin t}{\cos t} ∣ \underset{- \frac{π}{4}}{\overset{0}{}} = \sqrt{2} (0 - \ln (\sqrt{2} - 1)$ $= \sqrt{2} \ln (\frac{1}{\sqrt{2} - 1}) = \sqrt{2} \ln (\sqrt{2} + 1)$
	Answered by john santu last updated on 28/May/20
	$let x = (π/2)−t ⇒ { ((x=0 ,t=(π/2))),((x=(π/2),t=0)) :} ∫_(π/2) ^0 ((−dt)/(√(1+cos t))) = ∫_0 ^(π/2) (dt/(√(2cos^2 ((t/2))))) = (1/(√2)) ∫_0 ^(π/2) sec ((t/2)) dt = (√2) ∫_0 ^(π/2) sec ((t/2)) d((t/2)) = (√2) ln ∣sec ((t/2))+tan ((t/2))∣_0 ^(π/2) =(√(2 ()) ln((√2) + 1) −ln(1)) = (√2) ln ((√2) + 1 )$
	$let x = \frac{π}{2} - t \Rightarrow {\begin{cases} x = 0, t = \frac{π}{2} \\ x = \frac{π}{2}, t = 0 \end{cases}$ $\underset{\frac{π}{2}}{\int^{0}} \frac{- dt}{\sqrt{1 + \cos t}} = \underset{0}{\int^{\frac{π}{2}}} \frac{dt}{\sqrt{2 \cos^{2} (\frac{t}{2})}}$ $= \frac{1}{\sqrt{2}} \underset{0}{\int^{\frac{π}{2}}} \sec (\frac{t}{2}) dt = \sqrt{2} \underset{0}{\int^{\frac{π}{2}}} \sec (\frac{t}{2}) d (\frac{t}{2})$ $= \sqrt{2} \ln ∣ \sec (\frac{t}{2}) + \tan (\frac{t}{2}) ∣_{0}^{\frac{π}{2}}$ $= \sqrt{2 (} \ln (\sqrt{2} + 1) - \ln (1))$ $= \sqrt{2} \ln (\sqrt{2} + 1)$
	Answered by 1549442205 last updated on 28/May/20

	$putting \sqrt{1 + sinx} = t \Rightarrow t^{2} = 1 + sinx$ $\Rightarrow 2 tdt = cosxdx = \sqrt{1 - {(t^{2} - 1)}^{2}} dx$ $= t \sqrt{2 - t^{2}} dx . Hence, F (x) = \int \frac{2 dt}{t \sqrt{2 - t^{2}}}$ $putting \sqrt{2 - t^{2}} = u \Rightarrow \frac{- t}{\sqrt{2 - t^{2}}} dt = du \Rightarrow - tdt = udu$ $F = \int \frac{- 2 du}{2 - u^{2}} = \int \frac{2 du}{u^{2} - {(\sqrt{2})}^{2}} = \frac{1}{\sqrt{2}} \ln ∣ \frac{u - \sqrt{2}}{u + \sqrt{2}} ∣$ $\frac{1}{\sqrt{2}} \ln ∣ \frac{\sqrt{1 - sinx} - \sqrt{2}}{\sqrt{1 - \sin x} + \sqrt{2}} ∣ + C . Hence,$ $I = F (\frac{π}{2}) - F (0) = 0 - \frac{1}{\sqrt{2}} \ln \frac{\sqrt{2} - 1}{\sqrt{2} + 1}$ $= - \sqrt{2} \ln (\sqrt{2} - 1) = \sqrt{2} \ln (\sqrt{2} + 1)$
	Commented by john santu last updated on 28/May/20

	$oo you change it$
	Answered by mathmax by abdo last updated on 28/May/20
	$I =∫_0 ^(π/2) (dx/(√(1+sinx))) ⇒I =∫_0 ^(π/2) (dx/(√(cos^2 ((x/2))+sin^2 ((x/2)) +sin((x/2))cos((x/2))))) =∫_0 ^(π/2) (dx/(√((cos((x/2))+sin((x/2)))^2 ))) =∫_0 ^(π/2) (dx/(cos((x/2))+sin((x/2)))) =_((x/2)=t) ∫_0 ^(π/4) ((2dt)/(cost +sint)) =_(tan((t/2))=u) ∫_0 ^((√2)−1) ((4du)/((1+u^2 )(((1−u^2 )/(1+u^2 )) +((2u)/(1+u^2 ))))) =4 ∫_0 ^((√2)−1) (du/(1−u^2 +2u)) =−4 ∫_0 ^((√2)−1) (du/(u^2 −2u−1)) =−4 ∫_0 ^((√2)−1) (du/((u−1)^2 −2)) =−4 ∫_0 ^((√2)−1) (du/((u−1−(√2))(u−1+(√2)))) =((−4)/(2(√2)))∫_0 ^((√2)−1) {(1/(u−1−(√2)))−(1/(u−1+(√2)))}du =−(√2)[ln∣((u−1−(√2))/(u−1+(√2)))∣]_0 ^((√2)−1) =−(√2){ln∣((−2)/(2(√2)−2))∣−ln∣((1+(√2))/(−1+(√2)))∣} =−(√2){ln∣(1/(2−(√2)))∣−ln∣(((√2)+1)/((√2)−1))∣ =(√2){ln(2−(√2))+ln((((√2)+1)/((√2)−1)))}$
	$I = \int_{0}^{\frac{π}{2}} \frac{dx}{\sqrt{1 + sinx}} \Rightarrow I = \int_{0}^{\frac{π}{2}} \frac{dx}{\sqrt{\cos^{2} (\frac{x}{2}) + \sin^{2} (\frac{x}{2}) + \sin (\frac{x}{2}) \cos (\frac{x}{2})}}$ $= \int_{0}^{\frac{π}{2}} \frac{dx}{\sqrt{{(\cos (\frac{x}{2}) + \sin (\frac{x}{2}))}^{2}}} = \int_{0}^{\frac{π}{2}} \frac{dx}{\cos (\frac{x}{2}) + \sin (\frac{x}{2})} =_{\frac{x}{2} = t} \int_{0}^{\frac{π}{4}} \frac{2 dt}{cost + sint}$ $=_{\tan (\frac{t}{2}) = u} \int_{0}^{\sqrt{2} - 1} \frac{4 du}{(1 + u^{2}) (\frac{1 - u^{2}}{1 + u^{2}} + \frac{2 u}{1 + u^{2}})} = 4 \int_{0}^{\sqrt{2} - 1} \frac{du}{1 - u^{2} + 2 u}$ $= - 4 \int_{0}^{\sqrt{2} - 1} \frac{du}{u^{2} - 2 u - 1} = - 4 \int_{0}^{\sqrt{2} - 1} \frac{du}{{(u - 1)}^{2} - 2} = - 4 \int_{0}^{\sqrt{2} - 1} \frac{du}{(u - 1 - \sqrt{2}) (u - 1 + \sqrt{2})}$ $= \frac{- 4}{2 \sqrt{2}} \int_{0}^{\sqrt{2} - 1} {\frac{1}{u - 1 - \sqrt{2}} - \frac{1}{u - 1 + \sqrt{2}}} du$ $= - \sqrt{2} {[\ln ∣ \frac{u - 1 - \sqrt{2}}{u - 1 + \sqrt{2}} ∣]}_{0}^{\sqrt{2} - 1} = - \sqrt{2} {\ln ∣ \frac{- 2}{2 \sqrt{2} - 2} ∣ - \ln ∣ \frac{1 + \sqrt{2}}{- 1 + \sqrt{2}} ∣}$ $= - \sqrt{2} {\ln ∣ \frac{1}{2 - \sqrt{2}} ∣ - \ln ∣ \frac{\sqrt{2} + 1}{\sqrt{2} - 1} ∣ = \sqrt{2} {\ln (2 - \sqrt{2}) + \ln (\frac{\sqrt{2} + 1}{\sqrt{2} - 1})}$
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