Question Number 114196 by mathdave last updated on 17/Sep/20
$${solve} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{sin}{x}\right){dx} \\ $$
Commented by Dwaipayan Shikari last updated on 20/Sep/20
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}^{{n}} {x}}{{n}} \\ $$$$\overset{\infty} {\sum}\left(−\mathrm{1}\right)^{{n}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left({e}^{{ix}} −{e}^{−{ix}} \right)^{{n}} }{\mathrm{2}^{{n}} {i}^{{n}} {n}}{dx} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{{e}^{{inx}} }{\mathrm{2}^{{n}} .{i}^{{n}} {n}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{1}−{e}^{−\mathrm{2}{ix}} \right)^{{n}} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{n}} \frac{\mathrm{1}}{{n}}\:\frac{{e}^{{inx}} }{\mathrm{2}^{{n}} {i}^{{n}} }\left(\frac{\pi}{\mathrm{4}}+\frac{{n}}{\mathrm{2}{i}}{e}^{−\mathrm{2}{ix}} +…..\right) \\ $$
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...](https://www.tinkutara.com/question/Q114619.png)
$$\mathrm{let}\:\mathrm{take}\:\mathrm{a}\:\mathrm{try}\:\mathrm{with}\:\mathrm{this}\:\mathrm{integral} \\ $$$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{1}+\mathrm{cos}\left(\frac{\pi}{\mathrm{2}}−\mathrm{x}\right)\right)\mathrm{dx}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{2cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{4}}−\frac{\mathrm{x}}{\mathrm{2}}\right)\right)\mathrm{dx} \\ $$$$=\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}−\frac{\mathrm{x}}{\mathrm{2}}\right)\right)\mathrm{dx}\:\:\mathrm{changement}\:\frac{\pi}{\mathrm{4}}−\frac{\mathrm{x}}{\mathrm{2}}=\mathrm{t} \\ $$$$\mathrm{give}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\mathrm{cos}\left(\frac{\pi}{\mathrm{4}}−\frac{\mathrm{x}}{\mathrm{2}}\right)\right)\mathrm{dx}\:=\int_{\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{8}}} \mathrm{ln}\left(\mathrm{cost}\right)\left(−\mathrm{2dt}\right) \\ $$$$=\mathrm{2}\:\int_{\frac{\pi}{\mathrm{8}}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{ln}\left(\mathrm{cost}\right)\mathrm{dt}\:\:=\mathrm{2}\left\{\:\left[\mathrm{t}\:\mathrm{ln}\left(\mathrm{cost}\right)\right]_{\frac{\pi}{\mathrm{8}}} ^{\frac{\pi}{\mathrm{4}}} −\int_{\frac{\pi}{\mathrm{8}}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{t}×\frac{−\mathrm{sint}}{\mathrm{cost}}\:\mathrm{dt}\right\} \\ $$$$=\mathrm{2}\left\{\frac{\pi}{\mathrm{4}}\mathrm{ln}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)−\frac{\pi}{\mathrm{8}}\mathrm{ln}\left(\frac{\sqrt{\mathrm{2}+\sqrt{\mathrm{2}}}}{\mathrm{2}}\right)\:\right\}\:+\mathrm{2}\:\int_{\frac{\pi}{\mathrm{8}}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{t}\:\mathrm{tant}\:\mathrm{dt} \\ $$$$\int_{\frac{\pi}{\mathrm{8}}} ^{\frac{\pi}{\mathrm{4}}} \:\mathrm{t}\:\mathrm{tant}\:\mathrm{dt}\:=_{\mathrm{tant}\:=\mathrm{u}} \:\:\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\mathrm{1}} \:\frac{\mathrm{u}\:\mathrm{arctanu}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\:\mathrm{du} \\ $$$$=\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\mathrm{arctanu}\right]_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\mathrm{1}} −\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)×\frac{\mathrm{du}}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right)\frac{\pi}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{4}−\mathrm{2}\sqrt{\mathrm{2}}\right)×\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\:\mathrm{du} \\ $$$$\mathrm{but}\:\int_{\sqrt{\mathrm{2}}−\mathrm{1}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\mathrm{du}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\mathrm{du}−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\mathrm{du} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\mathrm{du}\:=_{\mathrm{u}=\mathrm{tan}\theta} \:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \theta}\right)}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \theta\right)\mathrm{d}\theta \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \mathrm{ln}\left(\mathrm{cos}\theta\right)\left(\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{this}\:\mathrm{integral}\:\mathrm{is}\:\mathrm{known}\:\mathrm{see}\:\mathrm{the}\right. \\ $$$$\left.\mathrm{platform}\:\mathrm{i}\:\mathrm{dont}\:\mathrm{remember}\:\mathrm{its}\:\mathrm{value}\right) \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \:\:\:\frac{\mathrm{ln}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\mathrm{du}\:=\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \mathrm{ln}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{u}^{\mathrm{2n}} \:\mathrm{du} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}−\mathrm{1}} \mathrm{u}^{\mathrm{2n}} \:\mathrm{ln}\left(\mathrm{1}+\mathrm{u}^{\mathrm{2}} \right)\mathrm{du}….\mathrm{be}\:\mathrm{continued}… \\ $$