solve ∫_0 ^(π/4) ln(1+sinx)dx


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Question Number 114196 by mathdave last updated on 17/Sep/20

$s o l v e$ $\int_{0}^{\frac{π}{4}} \ln (1 + \sin x) d x$
Commented by Dwaipayan Shikari last updated on 20/Sep/20

$\int_{0}^{\frac{π}{4}} {(- 1)}^{n} \underset{n = 1}{\sum^{\infty}} \frac{{s i n}^{n} x}{n}$ $\sum^{\infty} {(- 1)}^{n} \int_{0}^{\frac{π}{4}} \frac{{(e^{i x} - e^{- i x})}^{n}}{2^{n} i^{n} n} d x$ $\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} \frac{e^{i n x}}{2^{n} . i^{n} n} \int_{0}^{\frac{π}{4}} {(1 - e^{- 2 i x})}^{n}$ $\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n} \frac{1}{n} \frac{e^{i n x}}{2^{n} i^{n}} (\frac{π}{4} + \frac{n}{2 i} e^{- 2 i x} + . . . . .)$
	Answered by mathmax by abdo last updated on 19/Sep/20
	$let take a try with this integral I =∫_0 ^(π/4) ln(1+cos((π/2)−x))dx =∫_0 ^(π/4) ln(2cos^2 ((π/4)−(x/2)))dx =(π/4)ln(2)+2 ∫_0 ^(π/4) ln(cos((π/4)−(x/2)))dx changement (π/4)−(x/2)=t give ∫_0 ^(π/4) ln(cos((π/4)−(x/2)))dx =∫_(π/4) ^(π/8) ln(cost)(−2dt) =2 ∫_(π/8) ^(π/4) ln(cost)dt =2{ [t ln(cost)]_(π/8) ^(π/4) −∫_(π/8) ^(π/4) t×((−sint)/(cost)) dt} =2{(π/4)ln((1/(√2)))−(π/8)ln(((√(2+(√2)))/2)) } +2 ∫_(π/8) ^(π/4) t tant dt ∫_(π/8) ^(π/4) t tant dt =_(tant =u) ∫_((√2)−1) ^1 ((u arctanu)/(1+u^2 )) du =[(1/2)ln(1+u^2 )arctanu]_((√2)−1) ^1 −∫_((√2)−1) ^1 (1/2)ln(1+u^2 )×(du/(1+u^2 )) =(1/2)ln(2)(π/4)−(1/2)ln(4−2(√2))×(π/8)−(1/2) ∫_((√2)−1) ^1 ((ln(1+u^2 ))/(1+u^2 )) du but ∫_((√2)−1) ^1 ((ln(1+u^2 ))/(1+u^2 ))du =∫_0 ^1 ((ln(1+u^2 ))/(1+u^2 ))du−∫_0 ^((√2)−1) ((ln(1+u^2 ))/(1+u^2 ))du ∫_0 ^1 ((ln(1+u^2 ))/(1+u^2 ))du =_(u=tanθ) ∫_0 ^(π/4) ((ln((1/(cos^2 θ))))/(1+tan^2 θ))(1+tan^2 θ)dθ =−2∫_0 ^(π/4) ln(cosθ)(the value of this integral is known see the platform i dont remember its value) ∫_0 ^((√2)−1) ((ln(1+u^2 ))/(1+u^2 ))du =∫_0 ^((√2)−1) ln(1+u^2 )Σ_(n=0) ^∞ (−1)^n u^(2n) du =Σ_(n=0) ^∞ (−1)^n ∫_0 ^((√2)−1) u^(2n) ln(1+u^2 )du....be continued...$
	$let take a try with this integral$ $I = \int_{0}^{\frac{π}{4}} \ln (1 + \cos (\frac{π}{2} - x)) dx = \int_{0}^{\frac{π}{4}} \ln ({2 \cos}^{2} (\frac{π}{4} - \frac{x}{2})) dx$ $= \frac{π}{4} \ln (2) + 2 \int_{0}^{\frac{π}{4}} \ln (\cos (\frac{π}{4} - \frac{x}{2})) dx changement \frac{π}{4} - \frac{x}{2} = t$ $give \int_{0}^{\frac{π}{4}} \ln (\cos (\frac{π}{4} - \frac{x}{2})) dx = \int_{\frac{π}{4}}^{\frac{π}{8}} \ln (cost) (- 2 dt)$ $= 2 \int_{\frac{π}{8}}^{\frac{π}{4}} \ln (cost) dt = 2 {{[t \ln (cost)]}_{\frac{π}{8}}^{\frac{π}{4}} - \int_{\frac{π}{8}}^{\frac{π}{4}} t \times \frac{- sint}{cost} dt}$ $= 2 {\frac{π}{4} \ln (\frac{1}{\sqrt{2}}) - \frac{π}{8} \ln (\frac{\sqrt{2 + \sqrt{2}}}{2})} + 2 \int_{\frac{π}{8}}^{\frac{π}{4}} t tant dt$ $\int_{\frac{π}{8}}^{\frac{π}{4}} t tant dt =_{tant = u} \int_{\sqrt{2} - 1}^{1} \frac{u arctanu}{1 + u^{2}} du$ $= {[\frac{1}{2} \ln (1 + u^{2}) arctanu]}_{\sqrt{2} - 1}^{1} - \int_{\sqrt{2} - 1}^{1} \frac{1}{2} \ln (1 + u^{2}) \times \frac{du}{1 + u^{2}}$ $= \frac{1}{2} \ln (2) \frac{π}{4} - \frac{1}{2} \ln (4 - 2 \sqrt{2}) \times \frac{π}{8} - \frac{1}{2} \int_{\sqrt{2} - 1}^{1} \frac{\ln (1 + u^{2})}{1 + u^{2}} du$ $but \int_{\sqrt{2} - 1}^{1} \frac{\ln (1 + u^{2})}{1 + u^{2}} du = \int_{0}^{1} \frac{\ln (1 + u^{2})}{1 + u^{2}} du - \int_{0}^{\sqrt{2} - 1} \frac{\ln (1 + u^{2})}{1 + u^{2}} du$ $\int_{0}^{1} \frac{\ln (1 + u^{2})}{1 + u^{2}} du =_{u = \tan θ} \int_{0}^{\frac{π}{4}} \frac{\ln (\frac{1}{\cos^{2} θ})}{1 + \tan^{2} θ} (1 + \tan^{2} θ) d θ$ $= - 2 \int_{0}^{\frac{π}{4}} \ln (\cos θ) (the value of this integral is known see the$ $platform i dont remember its value)$ $\int_{0}^{\sqrt{2} - 1} \frac{\ln (1 + u^{2})}{1 + u^{2}} du = \int_{0}^{\sqrt{2} - 1} \ln (1 + u^{2}) \sum_{n = 0}^{\infty} {(- 1)}^{n} u^{2 n} du$ $= \sum_{n = 0}^{\infty} {(- 1)}^{n} \int_{0}^{\sqrt{2} - 1} u^{2 n} \ln (1 + u^{2}) du . . . . be continued . . .$
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