Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 52675 by maxmathsup by imad last updated on 11/Jan/19

$$let{u}_n={(-1)}^n{\int }_0^{\frac{\pi }{2}}{sin}^{n}xdxcalculate\mathrm{\Sigma }{u}_n$$

Commented by maxmathsup by imad last updated on 11/Jan/19

$$wehave\sum _{n=0}^{\mathrm{\infty }}{u}_n=\sum _{n=0}^{\mathrm{\infty }}{\int }_0^{\frac{\pi }{2}}{(-sinx)}^{n}dx={\int }_0^{\frac{\pi }{2}}\left(\sum _{n=0}^{\mathrm{\infty }}{(-sinx)}^n\right)dx$$$$={\int }_0^{\frac{\pi }{2}}\frac{dx}{1+sinx}{=}_{tan\left(\frac{x}{2}\right)=t}{\int }_0^1\frac{1}{1+\frac{2t}{1+t^2}}\frac{2dt}{1+t^2}=2{\int }_0^1\frac{dt}{1+t^2+2t}$$$$=2{\int }_0^1\frac{dt}{{(t+1)}^2}=2{[-\frac{1}{t+1}]}_0^1=2(1-\frac{1}{2})=1\Rightarrow \sum _{n=0}^{\mathrm{\infty }}{u}_n=1.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com