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Question Number 127528 by bramlexs22 last updated on 30/Dec/20

$${\int }_0^{\pi /4}\frac{\sin \mathrm{x}}{\sin \mathrm{x}+\cos \mathrm{x}}\mathrm{dx}=?$$

Answered by liberty last updated on 30/Dec/20

$$LetL={\int }_0^{\pi /4}\frac{\sin x}{\sin x+\cos x}dx$$$$andletY={\int }_0^{\pi /4}\frac{\cos x}{\sin x+\cos x}dx$$$$\left(1\right)L+Y={\int }_0^{\pi /4}\frac{\sin x+\cos x}{\sin x+\cos x}dx=\frac{\pi }{4}$$$$\left(2\right)L-Y={\int }_0^{\pi /4}\frac{\sin x-\cos x}{\sin x+\cos x}dx$$$$=-{\int }_0^{\pi /4}\frac{d(\sin x+\cos x)}{\sin x+\cos x}$$$$=-{\left[\ln (\sin x+\cos x)\right]}_0^{\pi /4}$$$$=-\frac{1}{2}\ln \left(2\right)$$$$\left(3\right)L=\frac{(L+Y)+(L-Y)}{2}=\frac{\pi }{8}-\frac{1}{4}\ln \left(2\right)$$$$$$

Commented by bramlexs22 last updated on 30/Dec/20

$$\mathrm{amazing}...$$

Answered by Dwaipayan Shikari last updated on 30/Dec/20

$${\int }_0^1\frac{t}{t+1}.\frac{1}{t^2+1}dttanx=t$$$$=\frac{1}{2}{\int }_0^{1}t(\frac{1}{t+1}-\frac{t-1}{t^2+1})dt$$$$=\frac{1}{2}{\int }_0^1-\frac{1}{t+1}+\frac{t}{t^2+1}+\frac{1}{1+t^2}dt={\left[\frac{1}{2}log\left(\frac{\sqrt{t^2+1}}{t+1}\right)\right]}_0^1+{\left[\frac{1}{2}{tan}^{-1}t\right]}_0^1$$$$=\frac{\pi }{8}-\frac{1}{4}log\left(2\right)$$

Answered by mathmax by abdo last updated on 31/Dec/20

$$\mathrm{I}={\int }_0^{\frac{\pi }{4}}\frac{\mathrm{sinx}}{\mathrm{sinx}+\mathrm{cosx}}\mathrm{dx}\mathrm{we}\mathrm{do}\mathrm{the}\mathrm{changement}\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}\Rightarrow$$$$\mathrm{I}={\int }_0^{\sqrt{2}-1}\frac{\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}}{\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}+\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}}\times \frac{2\mathrm{dt}}{1+{\mathrm{t}}^2}={\int }_0^{\sqrt{2}-1}\frac{4\mathrm{t}}{(1+{\mathrm{t}}^2)(-{\mathrm{t}}^2+2\mathrm{t}+1)}\mathrm{dt}$$$$=-4{\int }_0^{\sqrt{2}-1}\frac{\mathrm{tdt}}{({\mathrm{t}}^2+1)({\mathrm{t}}^2-2\mathrm{t}-1)}\mathrm{let}\mathrm{decompose}\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{t}}{({\mathrm{t}}^2+1)({\mathrm{t}}^2-2\mathrm{t}-1)}$$$${\mathrm{t}}^2-2\mathrm{t}-1=0\to {\mathrm{\Delta }}^{{}^{\prime }}=1+1=2\Rightarrow {\mathrm{t}}_1=1+\sqrt{2}\mathrm{and}{\mathrm{t}}_2=1-\sqrt{2}$$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{\mathrm{a}}{\mathrm{t}-{\mathrm{t}}_1}+\frac{\mathrm{b}}{\mathrm{t}-{\mathrm{t}}_2}+\frac{\mathrm{ct}+\mathrm{d}}{{\mathrm{t}}^2+1}$$$$\mathrm{a}=\frac{{\mathrm{t}}_1}{({\mathrm{t}}_1^2+1)({\mathrm{t}}_1-{\mathrm{t}}_2)}=\frac{1+\sqrt{2}}{2\sqrt{2}(1+3+2\sqrt{2})}=\frac{1+\sqrt{2}}{2\sqrt{2}(4+2\sqrt{2})}$$$$\mathrm{b}=\frac{{\mathrm{t}}_2}{({\mathrm{t}}_2^2+1)({\mathrm{t}}_2-{\mathrm{t}}_1)}=\frac{1-\sqrt{2}}{(-2\sqrt{2})(3-2\sqrt{2})+1)}=\frac{\sqrt{2}-1}{2\sqrt{2}(4-2\sqrt{2})}$$$${\lim }_{\mathrm{t}\to +\mathrm{\infty }}\mathrm{tF}\left(\mathrm{t}\right)=0=\mathrm{a}+\mathrm{b}+\mathrm{c}\Rightarrow \mathrm{c}=-\mathrm{a}-\mathrm{b}$$$$\mathrm{F}\left(0\right)=0=-\frac{\mathrm{a}}{{\mathrm{t}}_1}-\frac{\mathrm{b}}{{\mathrm{t}}_2}+\mathrm{d}\Rightarrow \mathrm{d}=\frac{\mathrm{a}}{{\mathrm{t}}_1}+\frac{\mathrm{b}}{{\mathrm{t}}_2}\Rightarrow$$$${\int }_0^{\sqrt{2}-1}\mathrm{F}\left(\mathrm{t}\right)\mathrm{dt}=\mathrm{a}{\int }_0^{\sqrt{2}-1}\frac{\mathrm{dt}}{\mathrm{t}-{\mathrm{t}}_1}+\mathrm{b}{\int }_0^{\sqrt{2}-1}\frac{\mathrm{dt}}{\mathrm{t}-{\mathrm{t}}_2}+\frac{\mathrm{c}}{2}{\int }_0^{\sqrt{2}-1}\frac{2\mathrm{t}}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$+\mathrm{d}{\int }_0^{\sqrt{2}-1}\frac{\mathrm{dt}}{{\mathrm{t}}^2+1}=\mathrm{a}{[\ln \mid \mathrm{t}-{\mathrm{t}}_1\mid ]}_0^{\sqrt{2}-1}+\mathrm{b}{[\ln \mid \mathrm{t}-{\mathrm{t}}_2\mid ]}_0^{\sqrt{2}-1}+\frac{\mathrm{c}}{2}{\left[\ln ({\mathrm{t}}^2+1)\right]}_0^{\sqrt{2}-1}$$$$+\mathrm{d}{\left[\mathrm{arctant}\right]}_0^{\sqrt{2}-1}\mathrm{rest}\mathrm{to}\mathrm{finish}\mathrm{calculus}...$$

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