Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 151220 by peter frank last updated on 19/Aug/21

$$\mathrm{show}\mathrm{that}$$$${\int }_0^{\frac{\pi }{2}}\frac{\cos \mathrm{x}}{\cos \mathrm{x}+\sin \mathrm{x}+1}\mathrm{dx}=\frac{1}{4}(\pi -2\ln 2)$$

Answered by ArielVyny last updated on 19/Aug/21

$$t=tan\left(\frac{x}{2}\right)\to dt=\frac{1}{2}(1+t^2)dx$$$${\int }_0^1\frac{\frac{1-t^2}{1+t^2}}{\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}+1}\times \frac{2}{1+t^2}dt=2{\int }_0^1\frac{\frac{1-t^2}{1+t^2}}{(1-t^2)+\left(2t\right)+(1+t^2)}dt$$$$=2{\int }_0^1\frac{1-t^2}{1+t^2}\times \frac{1}{2+2t}dt={\int }_0^1\frac{(1-t)}{(1+t^2)}dt={\int }_0^1\frac{1}{1+t^2}dt-{\int }_0^1\frac{t}{1+t^2}dt$$$$={\int }_0^1\frac{1}{1+t^2}dt-{\int }_0^1\frac{t}{1+t^2}dt=\frac{\pi }{4}-\mathrm{\Sigma }{(-1)}^n{\int }_0^1{t}^{2n+1}$$$$\frac{\pi }{4}-\mathrm{\Sigma }{(-1)}^n\frac{1}{2n+2}=\frac{\pi }{4}-\frac{1}{2}ln\left(2\right)$$$${\int }_0^{\frac{\pi }{2}}\frac{cosx}{cosx+sinx+1}=\frac{\pi }{4}-\frac{1}{2}ln(2+$$

Commented by peter frank last updated on 19/Aug/21

$$\mathrm{thank}\mathrm{you}$$

Answered by Lordose last updated on 19/Aug/21

$$$$$$\mathrm{\Omega }={\int }_0^{\frac{\mathit{\pi }}{2}}\frac{\cos \left(\mathrm{x}\right)}{\cos \left(\mathrm{x}\right)+\sin \left(\mathrm{x}\right)+1}\mathrm{dx}\stackrel{\mathrm{t}=\tan \left(\frac{\mathrm{x}}{2}\right)}{=}{\int }_0^1\frac{\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}}{(1+{\mathrm{t}}^2)(\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}+\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}+1)}\mathrm{dt}$$$$\mathrm{\Omega }=2{\int }_0^1\frac{(1-{\mathrm{t}}^2)}{(1+{\mathrm{t}}^2)(1-{\mathrm{t}}^2+2\mathrm{t}+1+{\mathrm{t}}^2)}={\int }_0^1\frac{1-\mathrm{t}}{1+{\mathrm{t}}^2}\mathrm{dt}$$$$\mathrm{\Omega }={\int }_0^1\frac{1}{1+{\mathrm{t}}^2}\mathrm{dt}-\frac{1}{2}{\int }_0^1\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}\mathrm{dt}$$$$\mathbf{\Omega }=\frac{\mathit{\pi }}{4}-\frac{1}{2}\mathbf{ln}\left(2\right)$$

Commented by peter frank last updated on 19/Aug/21

$$\mathrm{thank}\mathrm{you}$$

Answered by Lordose last updated on 19/Aug/21

$$$$$$\mathrm{\Omega }={\int }_0^{\frac{\mathit{\pi }}{2}}\frac{\cos \left(\mathrm{x}\right)}{\cos \left(\mathrm{x}\right)+\sin \left(\mathrm{x}\right)+1}\mathrm{dx}\stackrel{\mathrm{t}=\tan \left(\frac{\mathrm{x}}{2}\right)}{=}{\int }_0^1\frac{\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}}{(1+{\mathrm{t}}^2)(\frac{1-{\mathrm{t}}^2}{1+{\mathrm{t}}^2}+\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}+1)}\mathrm{dt}$$$$\mathrm{\Omega }=2{\int }_0^1\frac{(1-{\mathrm{t}}^2)}{(1+{\mathrm{t}}^2)(1-{\mathrm{t}}^2+2\mathrm{t}+1+{\mathrm{t}}^2)}={\int }_0^1\frac{1-\mathrm{t}}{1+{\mathrm{t}}^2}\mathrm{dt}$$$$\mathrm{\Omega }={\int }_0^1\frac{1}{1+{\mathrm{t}}^2}\mathrm{dt}-\frac{1}{2}{\int }_0^1\frac{2\mathrm{t}}{1+{\mathrm{t}}^2}\mathrm{dt}$$$$\mathbf{\Omega }=\frac{\mathit{\pi }}{4}-\frac{1}{2}\mathbf{ln}\left(2\right)$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com