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Question Number 102701 by 175mohamed last updated on 10/Jul/20

$$Evaluate:$$$$\int \frac{\sin x}{x}dx$$

Commented by  M±th+et+s last updated on 10/Jul/20

$$gotoQ.98713isolvedthisquestion$$

Answered by PRITHWISH SEN 2 last updated on 10/Jul/20

$$\int \frac{1}{\mathrm{x}}\{\mathrm{x}-\frac{{\mathrm{x}}^3}{3!}+\frac{{\mathrm{x}}^5}{5!}-...\}\mathrm{dx}$$$$=\underset{\mathrm{r}=1}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{r}+1}\frac{{\mathrm{x}}^{2\mathrm{r}-1}}{(2\mathrm{r}-1).(2\mathrm{r}-1)!}\mathbf{please}\mathbf{check}.$$

Answered by mathmax by abdo last updated on 10/Jul/20

$$\mathrm{at}\mathrm{form}\mathrm{of}\mathrm{serie}\mathrm{wehave}\mathrm{sinx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}+1}}{(2\mathrm{n}+1)!}\mathrm{with}\mathrm{radius}\mathrm{r}=+\mathrm{\infty }$$$$\Rightarrow \frac{\mathrm{sinx}}{\mathrm{x}}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}{\mathrm{x}}^{2\mathrm{n}}}{(2\mathrm{n}+1)!}\Rightarrow \int \frac{\mathrm{sinx}}{\mathrm{x}}\mathrm{dx}=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1)!}\int {\mathrm{x}}^{2\mathrm{n}}\mathrm{dx}$$$$=\sum _{\mathrm{n}=0}^{\mathrm{\infty }}\frac{{(-1)}^{\mathrm{n}}}{(2\mathrm{n}+1)!(2\mathrm{n}+1)}{\mathrm{x}}^{2\mathrm{n}+1}+\mathrm{C}$$$$$$

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