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 190841 by Rupesh123 last updated on 12/Apr/23

Answered by ARUNG_Brandon_MBU last updated on 12/Apr/23

$$I={\int }_0^{\pi }\frac{xdx}{1+\cos \alpha \sin x}={\int }_0^{\pi }\frac{\pi -x}{1+\cos \alpha \sin x}dx$$$$=\frac{1}{2}{\int }_0^{\pi }\frac{\pi }{1+\cos \alpha \sin x}dx=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{\pi }{1+\cos \alpha \frac{2t}{1+t^2}}\cdot \frac{2dt}{1+t^2}$$$$=\pi {\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+2t\cos \alpha +1}=\pi {\int }_0^{\mathrm{\infty }}\frac{dt}{{(t+\cos \alpha )}^2+{\sin }^2\alpha }$$$$=\frac{\pi }{\sin \alpha }{\left[\arctan \left(\frac{t+\cos \alpha }{\sin \alpha }\right)\right]}_0^{\mathrm{\infty }}=\frac{\pi }{\sin \alpha }(\frac{\pi }{2}-\arctan \left(\frac{1}{\tan \alpha }\right))$$$$=\frac{\pi }{\sin \alpha }\left(\arctan \left(\tan \alpha \right)\right)=\frac{\pi \alpha }{\sin \alpha }$$

Commented by Rupesh123 last updated on 12/Apr/23

Nice solution, sir!

Commented by mehdee42 last updated on 13/Apr/23

$$verygood$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com