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Question Number 102366 by bobhans last updated on 08/Jul/20

$$\underset{0}{\stackrel{\pi /2}{\int }}\frac{\cos x}{1+\cos x+\sin x}dx?$$

Answered by PRITHWISH SEN 2 last updated on 08/Jul/20

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{\sin \mathrm{x}}{1+\cos \mathrm{x}+\sin \mathrm{x}}\mathrm{dx}$$$$2\mathrm{I}={\int }_0^{\frac{\pi }{2}}\{1-\frac{1}{1+\cos \mathrm{x}+\sin \mathrm{x}}\}\mathrm{dx}$$$$=\frac{\pi }{2}-\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\frac{{\sec }^2\frac{\mathrm{x}}{2}}{1+\tan \frac{\mathrm{x}}{2}}\mathrm{dx}$$$$=\frac{\pi }{2}-\ln \mid 1+\tan \frac{\mathrm{x}}{2}{\mid }_0^{\frac{\pi }{2}}$$$$=\frac{\pi }{2}-\ln \mid 2\mid$$$$\therefore \mathbf{I}=\frac{\mathit{\pi }}{4}-\frac{1}{2}\mathbf{ln}2\mathbf{please}\mathbf{check}$$

Commented by Dwaipayan Shikari last updated on 08/Jul/20

$$Itisright$$

Commented by PRITHWISH SEN 2 last updated on 08/Jul/20

$$\mathrm{sorry}\mathrm{i}\mathrm{made}\mathrm{a}\mathrm{mistake}.\mathrm{thank}\mathrm{you}.$$

Commented by bobhans last updated on 09/Jul/20

$$yesthankyou$$

Answered by Dwaipayan Shikari last updated on 08/Jul/20

$${\int }_0^{\frac{\pi }{2}}\frac{cosx}{1+cosx+sinx}=\int \frac{sinx}{1+cosx+sinx}=I$$$$2I={\int }_0^{\frac{\pi }{2}}1-\frac{1}{1+cosx+sinx}dx$$$$2I=\frac{\pi }{2}-2{\int }_0^1\frac{1}{1+\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}}.\frac{1}{1+t^2}dtputtan\frac{x}{2}=t$$$$2I=\frac{\pi }{2}-2{\int }_0^1\frac{1}{2(1+t)}dt$$$$2I=\frac{\pi }{2}-{\left[log(1+t)\right]}_0^1$$$$2I=\frac{\pi }{2}-log2$$$$I=\frac{\pi }{4}-\frac{1}{2}log2$$

Answered by mathmax by abdo last updated on 08/Jul/20

$$\mathrm{I}={\int }_0^{\frac{\pi }{2}}\frac{\mathrm{cosx}}{1+\mathrm{cosx}+\mathrm{sinx}}\mathrm{dx}\mathrm{cha}7\mathrm{gement}\tan \left(\frac{\mathrm{x}}{2}\right)=\mathrm{t}\mathrm{give}$$$$\mathrm{I}={\int }_0^1\frac{\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}}{1+\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}+\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}}\times \frac{2\mathrm{dt}}{1+{\mathrm{t}}^2}=2{\int }_0^1\frac{1-{\mathrm{t}}^2}{(1+{\mathrm{t}}^2+1-{\mathrm{t}}^2+2\mathrm{t})({\mathrm{t}}^2+1)}\mathrm{dt}$$$$={\int }_0^1\frac{1-{\mathrm{t}}^2}{(\mathrm{t}+1)({\mathrm{t}}^2+1)}\mathrm{dt}={\int }_0^1\frac{1-\mathrm{t}}{{\mathrm{t}}^2+1}\mathrm{dt}={\int }_0^1\frac{\mathrm{dt}}{{\mathrm{t}}^2+1}-\frac{1}{2}{\int }_0^1\frac{2\mathrm{t}}{{\mathrm{t}}^2+1}\mathrm{dt}$$$$={\left[\arctan \left(\mathrm{t}\right)\right]}_0^1-\frac{1}{2}{\left[\ln ({\mathrm{t}}^2+1)\right]}_0^1=\frac{\pi }{4}-\frac{\ln \left(2\right)}{2}$$

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