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Question Number 184753 by cortano1 last updated on 11/Jan/23

Answered by qaz last updated on 11/Jan/23

$$\underset{r\to \mathrm{\infty }}{lim}\frac{r^c{\int }_0^{\pi /2}{x}^r\sin xdx}{{\int }_0^{\pi /2}{x}^r\cos xdx}=\underset{r\to \mathrm{\infty }}{lim}\frac{r^c{\int }_0^{\pi /2}{e}^{rln(\frac{\pi }{2}-x)}cosxdx}{{\int }_0^{\pi /2}{e}^{rln(\frac{\pi }{2}-x)}sinxdx}$$$$=\underset{r\to \mathrm{\infty }}{lim}\frac{r^c{\int }_0^{\pi /2}{e}^{rln(1-\frac{2}{\pi }x)}cosxdx}{{\int }_0^{\pi /2}{e}^{rln(1-\frac{2}{\pi }x)}sinxdx}=\underset{r\to \mathrm{\infty }}{lim}\frac{r^c{\int }_0^{\mathrm{\infty }}{e}^{-\frac{2}{\pi }rx}cosxdx}{{\int }_0^{\mathrm{\infty }}{e}^{-\frac{2}{\pi }rx}\sin xdx}.......bylaplacemethod$$$$=\underset{r\to \mathrm{\infty }}{lim}\frac{r^c\frac{\frac{2}{\pi }r}{{\left(\frac{2}{\pi }r\right)}^2+1}}{\frac{1}{{\left(\frac{2}{\pi }r\right)}^2+1}}=\underset{r\to \mathrm{\infty }}{lim}\frac{2r^{c+1}}{\pi }=L\Rightarrow c=-1,L=\frac{2}{\pi }$$

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