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Question Number 144815 by mnjuly1970 last updated on 29/Jun/21

Answered by mindispower last updated on 30/Jun/21

$$={\int }_0^{\mathrm{\infty }}\frac{sin(x+\sqrt{1+x^2})+sin(\sqrt{1+x^2}-x)}{2\sqrt{1+x^2}}dx$$$$x=sh\left(t\right)$$$${\int }_0^{\mathrm{\infty }}\frac{sin\left(e^t\right)+sin\left(e^{-t}\right)}{2ch\left(t\right)}ch\left(t\right)dt$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}sin\left(e^t\right)+sin\left(e^{-t}\right)dt,e^t=yfirt,e^{-t}=x2nd$$$$=\frac{1}{2}{\int }_1^{\mathrm{\infty }}\frac{sin\left(y\right)}{y}dx+\frac{1}{2}{\int }_0^1\frac{sin\left(x\right)}{x}dx$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{sin\left(t\right)}{t}dt=\frac{\pi }{4}$$

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