Question Number 46639 by Tinkutara last updated on 29/Oct/18
Commented by maxmathsup by imad last updated on 30/Oct/18
![let A =∫_0 ^π ((sin(2x))/(x+1))dx ⇒A=_(2x=t) ∫_0 ^(2π) ((sin(t))/((t/2)+1)) (dt/2) = ∫_0 ^(2π) ((sint)/(t+2)) =_(byparts) [−(1/(t+2)) cost]_0 ^(2π) −∫_0 ^(2π) ((−1)/((t+2)^2 )) (−cost)dt =(1/2) −(1/(2π +2)) −∫_0 ^(2π) ((cost)/((t+2)^2 )) dt for that i think that I = ∫_0 ^(2π) ((cosx)/((x+2)^2 ))dx ⇒ A =(1/2){1−(1/(π +1))}−I =(π/(2π+2)) −I .](https://www.tinkutara.com/question/Q46734.png)
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{sin}\left(\mathrm{2}{x}\right)}{{x}+\mathrm{1}}{dx}\:\:\:\Rightarrow{A}=_{\mathrm{2}{x}={t}} \:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sin}\left({t}\right)}{\frac{{t}}{\mathrm{2}}+\mathrm{1}}\:\frac{{dt}}{\mathrm{2}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{sint}}{{t}+\mathrm{2}}\:\:\:=_{{byparts}} \:\:\:\:\left[−\frac{\mathrm{1}}{{t}+\mathrm{2}}\:{cost}\right]_{\mathrm{0}} ^{\mathrm{2}\pi} \:−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{−\mathrm{1}}{\left({t}+\mathrm{2}\right)^{\mathrm{2}} }\:\left(−{cost}\right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:−\frac{\mathrm{1}}{\mathrm{2}\pi\:+\mathrm{2}}\:−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cost}}{\left({t}+\mathrm{2}\right)^{\mathrm{2}} }\:{dt}\:\:{for}\:{that}\:\:{i}\:{think}\:{that}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cosx}}{\left({x}+\mathrm{2}\right)^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\mathrm{1}−\frac{\mathrm{1}}{\pi\:+\mathrm{1}}\right\}−{I}\:=\frac{\pi}{\mathrm{2}\pi+\mathrm{2}}\:−{I}\:. \\ $$
Commented by Tinkutara last updated on 31/Oct/18
Thanks Sir! Maybe it's a mistake in question.
Answered by tanmay.chaudhury50@gmail.com last updated on 31/Oct/18
![∫((sin2x)/(x+1))dx (1/(x+1))∫sin2x−∫[(d/dx) ((1/(x+1)))∫sin2xdx]dx (1/(x+1))×((−cos2x)/2)−∫((−1)/((x+1)^2 ))×((−cos2x)/2)dx ((−cos2x)/(2(x+1)))−∫((cos2x)/(2(x+1)^2 ))dx so ∫_0 ^π ((sin2x)/(x+1))dx =∣((−cos2x)/(2(x+1)))∣_0 ^π −(1/2)∫_0 ^π ((cos2x)/((x+1)^2 ))dx =((−1)/2)(((cos2π)/(π+1))−((cos0)/(0+1)))−(1/2)j ((−1)/2)((1/(π+1))−1)−(1/2)j contd...](https://www.tinkutara.com/question/Q46669.png)
$$\int\frac{{sin}\mathrm{2}{x}}{{x}+\mathrm{1}}{dx} \\ $$$$\frac{\mathrm{1}}{{x}+\mathrm{1}}\int{sin}\mathrm{2}{x}−\int\left[\frac{{d}}{{dx}}\:\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)\int{sin}\mathrm{2}{xdx}\right]{dx} \\ $$$$\frac{\mathrm{1}}{{x}+\mathrm{1}}×\frac{−{cos}\mathrm{2}{x}}{\mathrm{2}}−\int\frac{−\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }×\frac{−{cos}\mathrm{2}{x}}{\mathrm{2}}{dx} \\ $$$$\frac{−{cos}\mathrm{2}{x}}{\mathrm{2}\left({x}+\mathrm{1}\right)}−\int\frac{{cos}\mathrm{2}{x}}{\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$${so}\:\int_{\mathrm{0}} ^{\pi} \frac{{sin}\mathrm{2}{x}}{{x}+\mathrm{1}}{dx} \\ $$$$=\mid\frac{−{cos}\mathrm{2}{x}}{\mathrm{2}\left({x}+\mathrm{1}\right)}\mid_{\mathrm{0}} ^{\pi} −\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi} \frac{{cos}\mathrm{2}{x}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{2}}\left(\frac{{cos}\mathrm{2}\pi}{\pi+\mathrm{1}}−\frac{{cos}\mathrm{0}}{\mathrm{0}+\mathrm{1}}\right)−\frac{\mathrm{1}}{\mathrm{2}}{j} \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\pi+\mathrm{1}}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}{j} \\ $$$${contd}… \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tinkutara last updated on 30/Oct/18
$${Why}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{cost}}{\left({t}+\mathrm{2}\right)^{\mathrm{2}} }{dt}=\mathrm{2}\int_{\mathrm{0}} ^{\pi} \frac{{cost}}{\left({t}+\mathrm{2}\right)^{\mathrm{2}} }{dt}=\mathrm{4}{I}? \\ $$$${Sir}\:{how}\:{f}\left(\mathrm{2}{a}−{x}\right)={f}\left({x}\right)\:{here}? \\ $$