Question Number 33599 by abdo imad last updated on 19/Apr/18
$${calculatef}\left({a}\right)=\:\:\int_{−{a}} ^{{a}} \:\:\:\:\frac{{dx}}{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\:{with}\:{a}>\mathrm{0}\:. \\ $$
Commented by abdo imad last updated on 20/Apr/18
![case 1 t≠o changement x=ttanθ give f(a) = ∫_(−arctan((a/t))) ^(arctan((a/t))) (1/((t^2 (1+tan^2 θ))^(3/2) )) t(1+tan^2 θ)dθ = ∫_(−arctan((a/t))) ^(arctan((a/t))) ((t(1+tan^2 θ))/(t^3 (1+tan^2 θ)^(3/2) ))dθ =(1/t^2 ) ∫_(−arctan((a/t))) ^(arctan((a/t))) (dθ/((1+tan^2 θ)^(1/2) )) =(2/t^2 ) ∫_0 ^(arctan((a/t))) cosθ dθ =(2/t^2 )[ sinθ]_0 ^(arcan((a/t))) = (2/t^2 ) sin(arctan((a/t))) but we have sin(arctanu)=(u/( (√(1+u^2 )))) f(a) = (2/t^2 ) ((a/t)/( (√(1+(a^2 /t^2 ))))) = ((2a)/t) (1/( (√((t^2 +a^2 )/t^2 )))) =((2a∣t∣)/(t(√(a^2 +t^2 )))) f(a) = ((2aξ(t))/( (√(t^2 +a^2 )))) with ξ(t)=1 if t>0 and ξ(t)=−1 if t<0 case 2 if t=0 f(a) = ∫_(−a) ^a (dx/x^3 ) =0 because x→x^3 is odd.](https://www.tinkutara.com/question/Q33632.png)
$${case}\:\mathrm{1}\:\:\:{t}\neq{o}\:\:{changement}\:{x}={ttan}\theta\:\:{give} \\ $$$${f}\left({a}\right)\:=\:\int_{−{arctan}\left(\frac{{a}}{{t}}\right)} ^{{arctan}\left(\frac{{a}}{{t}}\right)} \:\:\:\:\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} \left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:{t}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta \\ $$$$=\:\int_{−{arctan}\left(\frac{{a}}{{t}}\right)} ^{{arctan}\left(\frac{{a}}{{t}}\right)} \:\:\:\:\:\frac{{t}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)}{{t}^{\mathrm{3}} \left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{d}\theta \\ $$$$=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\int_{−{arctan}\left(\frac{{a}}{{t}}\right)} ^{{arctan}\left(\frac{{a}}{{t}}\right)} \:\:\:\:\:\:\:\frac{{d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{1}}{\mathrm{2}}} } \\ $$$$=\frac{\mathrm{2}}{{t}^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{{arctan}\left(\frac{{a}}{{t}}\right)} \:\:{cos}\theta\:{d}\theta\:\:=\frac{\mathrm{2}}{{t}^{\mathrm{2}} }\left[\:{sin}\theta\right]_{\mathrm{0}} ^{{arcan}\left(\frac{{a}}{{t}}\right)} \\ $$$$=\:\frac{\mathrm{2}}{{t}^{\mathrm{2}} }\:{sin}\left({arctan}\left(\frac{{a}}{{t}}\right)\right)\:\:{but}\:{we}\:{have}\:{sin}\left({arctanu}\right)=\frac{{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }} \\ $$$${f}\left({a}\right)\:=\:\frac{\mathrm{2}}{{t}^{\mathrm{2}} }\:\:\frac{\frac{{a}}{{t}}}{\:\sqrt{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{t}^{\mathrm{2}} }}}\:=\:\frac{\mathrm{2}{a}}{{t}}\:\:\:\frac{\mathrm{1}}{\:\sqrt{\frac{{t}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{t}^{\mathrm{2}} }}}\:\:=\frac{\mathrm{2}{a}\mid{t}\mid}{{t}\sqrt{{a}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }} \\ $$$${f}\left({a}\right)\:=\:\frac{\mathrm{2}{a}\xi\left({t}\right)}{\:\sqrt{{t}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }}\:\:\:{with}\:\:\xi\left({t}\right)=\mathrm{1}\:{if}\:{t}>\mathrm{0}\:{and}\:\xi\left({t}\right)=−\mathrm{1}\:{if}\:{t}<\mathrm{0} \\ $$$${case}\:\mathrm{2}\:{if}\:{t}=\mathrm{0} \\ $$$${f}\left({a}\right)\:=\:\int_{−{a}} ^{{a}} \:\:\frac{{dx}}{{x}^{\mathrm{3}} }\:=\mathrm{0}\:{because}\:{x}\rightarrow{x}^{\mathrm{3}} \:{is}\:{odd}. \\ $$
Commented by abdo imad last updated on 20/Apr/18
$${error}\:{at}\:{line}\:\mathrm{7}\:\:{f}\left({a}\right)\:=\:\frac{\mathrm{2}{a}}{{t}^{\mathrm{2}} }\:\frac{\mathrm{1}}{\:\sqrt{\frac{{t}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{{t}^{\mathrm{2}} }}}\:=\frac{\mathrm{2}{a}\:\mid{t}\mid}{{t}^{\mathrm{2}} \sqrt{{a}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }} \\ $$$$=\:\frac{\mathrm{2}{a}\xi\left({t}\right)}{{t}\sqrt{{a}^{\mathrm{2}} \:+{t}^{\mathrm{2}} }}\:. \\ $$
Answered by alex041103 last updated on 20/Apr/18
![∫(dx/((t^2 +x^2 )^(3/2) ))=(1/t^3 )∫(dx/((1+((x/t))^2 )^(3/2) )) Let u=x/t ⇒ dx=tdu (1/t^3 )∫(dx/((1+((x/t))^2 )^(3/2) ))=(1/t^2 )∫(du/((1+u^2 )^(3/2) )) We use the standart trigonometric substitution: u=tanθ du=sec^2 θdθ (1/t^2 )∫(du/((1+u^2 )^(3/2) ))=(1/t^2 )∫((sec^2 θdθ)/((1+tan^2 θ)^(3/2) )) And (1+tan^2 θ)^(3/2) =(sec^2 θ)^(3/2) =sec^3 θ ⇒(1/t^2 )∫((sec^2 θdθ)/((1+tan^2 θ)^(3/2) ))=(1/t^2 )∫((sec^2 θdθ)/(sec^3 θ))= =(1/t^2 )∫cosθdθ=(1/t^2 )sinθ u=tanθ⇒θ=arctan(u) ⇒(1/t^2 )sinθ=(1/t^2 ) (u/( (√(1+u^2 ))))=(u/(t(√(t^2 +(tu)^2 ))))= =(1/t^2 ) (x/( (√(t^2 +x^2 )))) ⇒∫_(−a) ^a (dx/((t^2 +x^2 )^(3/2) ))=(1/t^2 )[(x/( (√(t^2 +x^2 )))) ]_(−a) ^a = =((2a)/(t^2 (√(t^2 +a^2 )))) ⇒∫_(−a) ^a (dx/((t^2 +x^2 )^(3/2) ))=((2a)/(t^2 (√(t^2 +a^2 )))) Any questions?](https://www.tinkutara.com/question/Q33631.png)
$$\int\frac{{dx}}{\left({t}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }=\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\int\frac{{dx}}{\left(\mathrm{1}+\left(\frac{{x}}{{t}}\right)^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} } \\ $$$${Let}\:{u}={x}/{t}\:\Rightarrow\:{dx}={tdu} \\ $$$$\frac{\mathrm{1}}{{t}^{\mathrm{3}} }\int\frac{{dx}}{\left(\mathrm{1}+\left(\frac{{x}}{{t}}\right)^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\int\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} } \\ $$$${We}\:{use}\:{the}\:{standart}\:{trigonometric}\:{substitution}: \\ $$$${u}={tan}\theta\:{du}={sec}^{\mathrm{2}} \theta{d}\theta \\ $$$$\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\int\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\int\frac{{sec}^{\mathrm{2}} \theta{d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{3}/\mathrm{2}} } \\ $$$${And}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{3}/\mathrm{2}} =\left({sec}^{\mathrm{2}} \theta\right)^{\mathrm{3}/\mathrm{2}} ={sec}^{\mathrm{3}} \theta \\ $$$$\Rightarrow\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\int\frac{{sec}^{\mathrm{2}} \theta{d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{3}/\mathrm{2}} }=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\int\frac{{sec}^{\mathrm{2}} \theta{d}\theta}{{sec}^{\mathrm{3}} \theta}= \\ $$$$=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\int{cos}\theta{d}\theta=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{sin}\theta \\ $$$${u}={tan}\theta\Rightarrow\theta={arctan}\left({u}\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{sin}\theta=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\frac{{u}}{\:\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }}=\frac{{u}}{{t}\sqrt{{t}^{\mathrm{2}} +\left({tu}\right)^{\mathrm{2}} }}= \\ $$$$=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\:\frac{{x}}{\:\sqrt{{t}^{\mathrm{2}} +{x}^{\mathrm{2}} }} \\ $$$$\Rightarrow\int_{−{a}} ^{{a}} \:\frac{{dx}}{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }=\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\left[\frac{{x}}{\:\sqrt{{t}^{\mathrm{2}} +{x}^{\mathrm{2}} }}\:\right]_{−{a}} ^{{a}} = \\ $$$$=\frac{\mathrm{2}{a}}{{t}^{\mathrm{2}} \sqrt{{t}^{\mathrm{2}} +{a}^{\mathrm{2}} }} \\ $$$$\Rightarrow\int_{−{a}} ^{{a}} \:\frac{{dx}}{\left({t}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }=\frac{\mathrm{2}{a}}{{t}^{\mathrm{2}} \sqrt{{t}^{\mathrm{2}} +{a}^{\mathrm{2}} }} \\ $$$${Any}\:{questions}? \\ $$
Commented by abdo imad last updated on 20/Apr/18
$${you}\:{have}\:{commited}\:{a}\:{error}\:{sir}\:{alex}\:. \\ $$
Commented by alex041103 last updated on 20/May/18
$${Can}\:{you}\:{point}\:{it}\:{out},\:{so}\:{I}\:{can}\:{fix}\:{it}. \\ $$