Question Number 40157 by maxmathsup by imad last updated on 16/Jul/18
$${find}\:{the}\:{value}\:{of}\:\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:−\mathrm{2}{t}\:+\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$
Commented by maxmathsup by imad last updated on 16/Jul/18
$${let}\:{I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dt}}{\left({t}^{\mathrm{2}} \:−\mathrm{2}{t}\:+\mathrm{2}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$${I}\:=\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dt}}{\left\{\:\left({t}−\mathrm{1}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:{changement}\:{t}−\mathrm{1}={tanx}\:{give} \\ $$$${I}\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right){dx}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:=\:\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {x}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\:{dx} \\ $$$$=\int_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:{cosxdx}\:\:=\left[{sinx}\right]_{−\frac{\pi}{\mathrm{2}}} ^{\frac{\pi}{\mathrm{2}}} \:\:=\mathrm{2} \\ $$$${I}\:=\mathrm{2}\: \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jul/18
$$\int_{−\infty} ^{+\infty} \frac{{dt}}{\left\{\left({t}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}\right\}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:{let}\:\left({t}−\mathrm{1}\right)={tank} \\ $$$$\int_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} \frac{{sec}^{\mathrm{2}} {k}\:{dk}}{{sec}^{\mathrm{3}} {k}} \\ $$$$\int_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} {cosk}\:{dk} \\ $$$$\mid{sink}\mid_{−\frac{\Pi}{\mathrm{2}}} ^{\frac{\Pi}{\mathrm{2}}} =\mathrm{2} \\ $$