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Question Number 32349 by abdo imad last updated on 23/Mar/18

$$findthevalueof{\int }_0^{\pi }\frac{xdx}{1+sinx}.$$

Commented by abdo imad last updated on 24/Mar/18

$$thech.tan\left(\frac{x}{2}\right)=tgive$$$$I={\int }_0^{\pi }\frac{xdx}{1+sinx}={\int }_0^{\mathrm{\infty }}\frac{2artant}{1+\frac{2t}{1+t^2}}\frac{2dt}{1+t^2}$$$$=4{\int }_0^{\mathrm{\infty }}\frac{arctant}{1+t^2+2t}dt=4{\int }_0^{\mathrm{\infty }}\frac{arctant}{{(t+1)}^2}dt{}^{\prime }bypartsweget$$$$I=4({[-\frac{1}{t+1}arctant]}_o^{+\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}-\frac{1}{t+1}\frac{dt}{1+t^2})$$$$I=4{\int }_0^{\mathrm{\infty }}\frac{dt}{(t+1)(t^2+1)}letdecompose$$$$f\left(t\right)=\frac{1}{(t+1)(t^2+1)}=\frac{a}{t+1}+\frac{bt+c}{t^2+1}$$$$a={lim}_{t\to -1}(t+1)f\left(t\right)=\frac{1}{2}$$$${lim}_{t\to +\mathrm{\infty }}tf\left(t\right)=0=a+b\Rightarrow b=-a=-\frac{1}{2}\Rightarrow$$$$f\left(t\right)=\frac{1}{2(t+1)}+\frac{\frac{-1}{2}t+c}{t^2+1}$$$$f\left(0\right)=1=\frac{1}{2}+c\Rightarrow c=\frac{1}{2}\Rightarrow f\left(t\right)=\frac{1}{2(t+1)}+\frac{1}{2}\frac{-t+1}{t^2+1}$$$$I=2{\int }_0^{\mathrm{\infty }}(\frac{1}{t+1}-\frac{t-1}{t^2+1})dt$$$$=2{\int }_0^{\mathrm{\infty }}(\frac{1}{t+1}-\frac{1}{2}\frac{2t}{t^2+1})dt+2{\int }_0^{\mathrm{\infty }}\frac{dt}{t^2+1}$$$$I=2{[ln\mid t+1\mid -ln\sqrt{t^2+1}]}_0^{\mathrm{\infty }}+\pi$$$$I=2{[ln\mid \frac{t+1}{\sqrt{t^2+1}}\mid ]}_0^{\mathrm{\infty }}+\pi =0+\pi \Rightarrow I=\pi .$$

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