Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 127888 by A8;15: last updated on 02/Jan/21

Answered by mindispower last updated on 02/Jan/21

$$x=sh\left(t\right)$$$$\Rightarrow dx=ch\left(t\right)dt$$$$\iff {\int }_0^{\mathrm{\infty }}sin\left(ch\left(t\right)\right)cos\left(sh\left(t\right)\right)dt..=\mathrm{\Omega }$$$$\mathrm{\Lambda }={\int }_0^{\mathrm{\infty }}cos\left(ch\left(t\right)\right)sin\left(sh\left(t\right)\right)dt$$$${\int }_0^{\mathrm{\infty }}sin\left(e^t\right)dt={\int }_0^{\mathrm{\infty }}\frac{sin\left(x\right)}{x}dx=\frac{\pi }{2}$$$${\int }_0^{\mathrm{\infty }}sin(ch\left(t\right)+sh\left(t\right))dt={\int }_0^{\mathrm{\infty }}sin\left(e^t\right)dt={\int }_1^{\mathrm{\infty }}\frac{sin\left(x\right)}{x}dx$$$$\mathrm{\Delta }={\int }_0^{\mathrm{\infty }}(sin\left(ch\left(t\right)\right)cos\left(sh\left(t\right)\right)+cos\left(ch\left(t\right)\right)sin\left(sh\left(t\right)\right))dt$$$$\mathrm{\Omega }+\mathrm{\Lambda }={\int }_1^{\mathrm{\infty }}\frac{sin\left(x\right)}{t}dt$$$$\mathrm{\Omega }-\mathrm{\Lambda }={\int }_0^{\mathrm{\infty }}sin\left(e^{-t}\right)={\int }_0^1\frac{sin\left(t\right)}{t}dt$$$$\Rightarrow 2\mathrm{\Omega }={\int }_0^1\frac{sin\left(t\right)}{t}dt+{\int }_1^{\mathrm{\infty }}\frac{sin\left(t\right)}{t}dt={\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)}{t}dt=\frac{\pi }{2}$$$$\iff \mathrm{\Omega }=\frac{\pi }{4}$$$$$$$$$$$$$$

Commented by A8;15: last updated on 03/Jan/21

$$\mathrm{thanks}\mathrm{sir}.\mathbf{Happy}\mathbf{New}\mathbf{Year}$$

Commented by mindispower last updated on 07/Jan/21

$$withepleasur$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com