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Question Number 42793 by maxmathsup by imad last updated on 02/Sep/18

find f(a)= ∫_0 ^1 (dt/((a^2 +t^2 )^3 )) with a>0

$${find}\:{f}\left({a}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left({a}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:\:{with}\:{a}>\mathrm{0} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18

t=atanα dt=asec^2 αdα ∫_0 ^(tan^(−1) ((1/a))) ((asec^2 αdα)/({a^2 (1+tan^2 α)}^3 )) ∫_(0 ) ^(tan^(−1) ((1/a))) ((asec^2 αdα)/(a^6 sec^6 α)) (1/a^5 )∫_0 ^(tan^(−1) ((1/a))) (((1+cos2α)/2))^2 dα (1/(4a^5 ))∫_0 ^(tan^(−1) ((1/a))) 1+2cos2α+((1+cos4α)/2) dα (1/(8a^5 ))∫_0 ^(tan^(−1) ((1/a))) 3+4cos2α+cos4α dα (1/(8a^5 ))∣(3α+((4sin2α)/2)+((sin4α)/4))∣_0 ^(tan^(−1) ((1/a))) now tanα=(1/a) sin2α=((2tanα)/(1+tan^2 α))=((2/a)/(1+(1/a^2 )))=((2a)/(1+a^2 )) sin4α=2sin2α.cos2α =2(((2t)/(1+t^2 ))×((1−t^2 )/(1+t^2 ))) =((4((1/a))(1−(1/a^2 )))/((1+(1/a^2 ))^2 ))=(((4(a^2 −1))/a^3 )/(((a^2 +1)^2 )/a^4 ))=((4a(a^2 −1))/((a^2 +1)^2 )) now put the value... contd so the ans is =(1/(8a^5 ))∣3tan^(−1) ((1/a))+2.((2a)/(1+a^2 ))+(1/4).((4a(a^2 −1))/((a^2 +1)^2 ))∣ =(1/(8a^5 )){3tan^(−1) ((1/a))+((4a)/(1+a^2 ))+((a^3 −a)/((a^2 +1)^2 ))}

$${t}={atan}\alpha\:\:{dt}={asec}^{\mathrm{2}} \alpha{d}\alpha \\ $$ $$\int_{\mathrm{0}} ^{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{a}}\right)} \frac{{asec}^{\mathrm{2}} \alpha{d}\alpha}{\left\{{a}^{\mathrm{2}} \left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)\right\}^{\mathrm{3}} } \\ $$ $$\int_{\mathrm{0}\:} ^{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{a}}\right)} \frac{{asec}^{\mathrm{2}} \alpha{d}\alpha}{{a}^{\mathrm{6}} {sec}^{\mathrm{6}} \alpha} \\ $$ $$\frac{\mathrm{1}}{{a}^{\mathrm{5}} }\int_{\mathrm{0}} ^{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{a}}\right)} \left(\frac{\mathrm{1}+{cos}\mathrm{2}\alpha}{\mathrm{2}}\right)^{\mathrm{2}} {d}\alpha \\ $$ $$\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{5}} }\int_{\mathrm{0}} ^{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{a}}\right)} \mathrm{1}+\mathrm{2}{cos}\mathrm{2}\alpha+\frac{\mathrm{1}+{cos}\mathrm{4}\alpha}{\mathrm{2}}\:{d}\alpha \\ $$ $$\frac{\mathrm{1}}{\mathrm{8}{a}^{\mathrm{5}} }\int_{\mathrm{0}} ^{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{a}}\right)} \:\:\mathrm{3}+\mathrm{4}{cos}\mathrm{2}\alpha+{cos}\mathrm{4}\alpha\:{d}\alpha \\ $$ $$\frac{\mathrm{1}}{\mathrm{8}{a}^{\mathrm{5}} }\mid\left(\mathrm{3}\alpha+\frac{\mathrm{4}{sin}\mathrm{2}\alpha}{\mathrm{2}}+\frac{{sin}\mathrm{4}\alpha}{\mathrm{4}}\right)\mid_{\mathrm{0}} ^{{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{a}}\right)} \\ $$ $${now}\:{tan}\alpha=\frac{\mathrm{1}}{{a}} \\ $$ $${sin}\mathrm{2}\alpha=\frac{\mathrm{2}{tan}\alpha}{\mathrm{1}+{tan}^{\mathrm{2}} \alpha}=\frac{\frac{\mathrm{2}}{{a}}}{\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }}=\frac{\mathrm{2}{a}}{\mathrm{1}+{a}^{\mathrm{2}} } \\ $$ $${sin}\mathrm{4}\alpha=\mathrm{2}{sin}\mathrm{2}\alpha.{cos}\mathrm{2}\alpha \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\left(\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{2}} }×\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right) \\ $$ $$\:\:=\frac{\mathrm{4}\left(\frac{\mathrm{1}}{{a}}\right)\left(\mathrm{1}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)}{\left(\mathrm{1}+\frac{\mathrm{1}}{{a}^{\mathrm{2}} }\right)^{\mathrm{2}} }=\frac{\frac{\mathrm{4}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}{{a}^{\mathrm{3}} }}{\frac{\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }{{a}^{\mathrm{4}} }}=\frac{\mathrm{4}{a}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$ $${now}\:{put}\:{the}\:{value}... \\ $$ $${contd}\:\:{so}\:{the}\:{ans}\:{is} \\ $$ $$=\frac{\mathrm{1}}{\mathrm{8}{a}^{\mathrm{5}} }\mid\mathrm{3}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{a}}\right)+\mathrm{2}.\frac{\mathrm{2}{a}}{\mathrm{1}+{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}}.\frac{\mathrm{4}{a}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\mid \\ $$ $$=\frac{\mathrm{1}}{\mathrm{8}{a}^{\mathrm{5}} }\left\{\mathrm{3}{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{{a}}\right)+\frac{\mathrm{4}{a}}{\mathrm{1}+{a}^{\mathrm{2}} }+\frac{{a}^{\mathrm{3}} −{a}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\right\} \\ $$

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