Question Number 53293 by gunawan last updated on 20/Jan/19
$$\int_{−\mathrm{1}/\mathrm{2}} ^{\mathrm{1}/\mathrm{2}} \mid{x}\mathrm{cos}\:\frac{\pi{x}}{\mathrm{2}}\mid\:{dx}=… \\ $$
Commented by maxmathsup by imad last updated on 20/Jan/19
$${we}\:{have}\:{f}\left({x}\right)=\mid{xcos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\mid=\mid{x}\mid\mid{cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\mid\:{is}\:{a}\:{even}\:{function}\:{so} \\ $$$${I}=\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {f}\left({x}\right){dx}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {x}\mid{cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\mid\:{dx}\:\:{but}\:\:\mathrm{0}\leqslant{x}\leqslant\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{0}\leqslant\pi{x}\leqslant\frac{\pi}{\mathrm{2}}\:\Rightarrow\mathrm{0}\leqslant\frac{\pi{x}}{\mathrm{2}}\leqslant\frac{\pi}{\mathrm{4}} \\ $$$$\Rightarrow\:{cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\geqslant\mathrm{0}\:\Rightarrow\:{I}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:{x}\:{cos}\left(\frac{\pi{x}}{\mathrm{2}}\right){dx}\:{by}\:{parts}\:{u}={x}\:{and}\:{v}^{'} ={cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\:\Rightarrow \\ $$$${I}\:=\mathrm{2}\left\{\:\:\left[\frac{\mathrm{2}}{\pi}{sin}\left(\frac{\pi{x}}{\mathrm{2}}\right){x}\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:−\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\frac{\mathrm{2}}{\pi}{sin}\left(\frac{\pi{x}}{\mathrm{2}}\right){dx}\right\} \\ $$$$=\mathrm{2}\left\{\frac{\mathrm{1}}{\pi}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right\}−\frac{\mathrm{4}}{\pi}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:{sin}\left(\frac{\pi{x}}{\mathrm{2}}\right){dx}\:=\frac{\sqrt{\mathrm{2}}}{\pi}\:+\frac{\mathrm{4}}{\pi}\:\left[\frac{\mathrm{2}}{\pi}{cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\right]_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\pi}\:+\frac{\mathrm{8}}{\pi^{\mathrm{2}} }\left(\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:−\mathrm{1}\right)\:=\frac{\sqrt{\mathrm{2}}}{\pi}\:\:+\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\left(\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{8}\right)\:=\frac{\pi\sqrt{\mathrm{2}}+\mathrm{4}\sqrt{\mathrm{2}}−\mathrm{8}}{\pi^{\mathrm{2}} }\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19
$${f}\left({x}\right)={xcos}\left(\frac{\pi{x}}{\mathrm{2}}\right) \\ $$$${f}\left(−{x}\right)=\left(−{x}\right){cos}\left(\frac{−\pi{x}}{\mathrm{2}}\right)=−{xcos}\left(\frac{\pi{x}}{\mathrm{2}}\right) \\ $$$${but}\:\mid−{xcos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\mid={xcos}\left(\frac{\pi{x}}{\mathrm{2}}\right) \\ $$$$\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\mid\frac{{xcos}\left(\frac{\pi{x}}{\mathrm{2}}\right)}{}\mid=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \mid{xcos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\mid \\ $$$$\mathrm{2}×\int_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} {xcos}\left(\frac{\pi{x}}{\mathrm{2}}\right){dx} \\ $$$${I}_{\mathrm{1}} =\int{xcos}\left(\frac{\pi{x}}{\mathrm{2}}\right){dx} \\ $$$$={x}×\frac{{sin}\left(\frac{\pi{x}}{\mathrm{2}}\right)}{\frac{\pi}{\mathrm{2}}}−\mathrm{1}×\int\frac{{sin}\left(\frac{\pi{x}}{\mathrm{2}}\right)}{\frac{\pi}{\mathrm{2}}}{dx} \\ $$$$=\frac{\mathrm{2}{x}}{\pi}{sin}\left(\frac{\pi{x}}{\mathrm{2}}\right)+\frac{\mathrm{2}}{\pi}×\frac{{cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)}{\frac{\pi}{\mathrm{2}}} \\ $$$$=\frac{\mathrm{2}{x}}{\pi}{sin}\left(\frac{\pi{x}}{\mathrm{2}}\right)+\frac{\mathrm{4}}{\pi^{\mathrm{2}} }{cos}\left(\frac{\pi{x}}{\mathrm{2}}\right) \\ $$$${required}\:{answer}\:{is} \\ $$$$\mathrm{2}×\mid\frac{\mathrm{2}{x}}{\pi}{sin}\left(\frac{\pi{x}}{\mathrm{2}}\right)+\frac{\mathrm{4}}{\pi^{\mathrm{2}} }{cos}\left(\frac{\pi{x}}{\mathrm{2}}\right)\mid_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$=\mathrm{2}×\left[\left\{\frac{\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}}{\pi}{sin}\left(\frac{\pi}{\mathrm{4}}\right)+\frac{\mathrm{4}}{\pi^{\mathrm{2}} }{cos}\left(\frac{\pi}{\mathrm{4}}\right)\right\}−\left\{\mathrm{0}+\frac{\mathrm{4}}{\pi^{\mathrm{2}} }{cos}\mathrm{0}\right\}\right] \\ $$$$=\mathrm{2}×\left[\frac{\mathrm{1}}{\pi\sqrt{\mathrm{2}}}+\frac{\mathrm{4}}{\pi^{\mathrm{2}} \sqrt{\mathrm{2}}}−\frac{\mathrm{4}}{\pi^{\mathrm{2}} }\right] \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\pi}+\frac{\mathrm{8}}{\pi^{\mathrm{2}} \sqrt{\mathrm{2}}}−\frac{\mathrm{8}}{\pi^{\mathrm{2}} }\:{pls}\:{check}\:{the}\:{steps}\:{mistake}\:{if}\:{any} \\ $$
Commented by gunawan last updated on 20/Jan/19
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