Question Number 111023 by mohammad17 last updated on 01/Sep/20
$$\int\frac{{sin}\left({x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$
Commented by mohammad17 last updated on 01/Sep/20
$${help}\:{me}\:{sir} \\ $$
Answered by mathdave last updated on 02/Sep/20
![solution series form of (1/(1+x^2 ))=(−1)^n Σ_(n=0) ^∞ x^(2n) I=∫((sinx)/(1+x^2 ))dx=(−1)^n Σ_(n=0) ^∞ ∫x^(2n) sinxdx using IBP let u=x^(2n) ,du=2x^(2n) and ∫dv=∫sinxdx,v=−cosx but ∫udv=uv−∫vdu I=(−1)^n Σ_(n=0) ^∞ [(−x^(2n) cosx+2∫x^(2n) cosxdx] I=−(−1)^n Σ_(n=0) ^∞ x^(2n) cosx+2(−1)^n Σ_(n=0) ^∞ ∫x^(2n) cosxdx I=−(−1)^n Σ_(n=0) ^∞ x^(2n) cosx+2(−1)^n Σ_(n=0) ^∞ [x^(2n) sinx−2∫x^(2n) sinxdx] I=−(−1)^n Σ_(n=0) ^∞ x^(2n) cosx+2(−1)^n Σ_(n=0) ^∞ x^(2n) sinx−4I I+4I=(−1)^n Σ_(n=0) ^∞ x^(2n) [2sinx−cosx] ∵I=∫((sinx)/(1+x^2 ))dx=(((−1)^n Σ_(n=0) ^∞ x^(2n) )/5)[2sinx−cosx] mathdave(02/09/2020)](https://www.tinkutara.com/question/Q111136.png)
$${solution}\: \\ $$$${series}\:{form}\:{of}\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }=\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} \\ $$$${I}=\int\frac{\mathrm{sin}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int{x}^{\mathrm{2}{n}} \mathrm{sin}{xdx} \\ $$$${using}\:{IBP} \\ $$$${let}\:{u}={x}^{\mathrm{2}{n}} ,{du}=\mathrm{2}{x}^{\mathrm{2}{n}} {and}\:\:\int{dv}=\int\mathrm{sin}{xdx},{v}=−\mathrm{cos}{x} \\ $$$${but}\:\int{udv}={uv}−\int{vdu} \\ $$$${I}=\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[\left(−{x}^{\mathrm{2}{n}} \mathrm{cos}{x}+\mathrm{2}\int{x}^{\mathrm{2}{n}} \mathrm{cos}{xdx}\right]\right. \\ $$$${I}=−\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} \mathrm{cos}{x}+\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\int{x}^{\mathrm{2}{n}} \mathrm{cos}{xdx} \\ $$$${I}=−\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} \mathrm{cos}{x}+\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left[{x}^{\mathrm{2}{n}} \mathrm{sin}{x}−\mathrm{2}\int{x}^{\mathrm{2}{n}} \mathrm{sin}{xdx}\right] \\ $$$${I}=−\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} \mathrm{cos}{x}+\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} \mathrm{sin}{x}−\mathrm{4}{I} \\ $$$${I}+\mathrm{4}{I}=\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} \left[\mathrm{2sin}{x}−\mathrm{cos}{x}\right] \\ $$$$\because{I}=\int\frac{\mathrm{sin}{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\left(−\mathrm{1}\right)^{{n}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{n}} }{\mathrm{5}}\left[\mathrm{2sin}{x}−\mathrm{cos}{x}\right] \\ $$$${mathdave}\left(\mathrm{02}/\mathrm{09}/\mathrm{2020}\right) \\ $$
Commented by mohammad17 last updated on 02/Sep/20
$${you}\:{are}\:{an}\:{awesome}\:{person}\:{thank}\:{you}\:{sir} \\ $$