Question Number 66328 by mathmax by abdo last updated on 12/Aug/19
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} } \\ $$
Commented by mathmax by abdo last updated on 15/Aug/19
![let I =∫_0 ^1 (dt/((1+t^2 )^3 )) changement t=tanθ give I =∫_0 ^(π/4) ((1+tan^2 θ)/((1+tan^2 θ)^3 ))dθ =∫_0 ^(π/4) (dθ/((1+tan^2 θ)^2 )) =∫_0 ^(π/4) cos^4 θdθ =∫_0 ^(π/4) (((1+cos(2θ))/2))^2 dθ =(1/4)∫_0 ^(π/4) (1+2cos(2θ) +((1+cos(4θ))/2))dθ =(1/8)∫_0 ^(π/4) {3 +4cos(2θ)+cos(4θ)}dθ =(3/8)(π/4) +(1/2) ∫_0 ^(π/4) cos(2θ)dθ +(1/8)∫_0 ^(π/4) cos(4θ)dθ =((3π)/(32)) +(1/4)[sin(2θ)]_0 ^(π/4) +(1/(32))[sin(4θ)]_0 ^(π/4) =((3π)/(32)) +(1/4) +0 ⇒ I =((3π)/(32)) +(1/4)](https://www.tinkutara.com/question/Q66458.png)
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{3}} }\:{changement}\:{t}={tan}\theta\:{give} \\ $$$${I}\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{\mathrm{1}+{tan}^{\mathrm{2}} \theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{3}} }{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{{d}\theta}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}^{\mathrm{4}} \theta{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\frac{\mathrm{1}+{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}\right)^{\mathrm{2}} {d}\theta\:=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left(\mathrm{1}+\mathrm{2}{cos}\left(\mathrm{2}\theta\right)\:+\frac{\mathrm{1}+{cos}\left(\mathrm{4}\theta\right)}{\mathrm{2}}\right){d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \left\{\mathrm{3}\:+\mathrm{4}{cos}\left(\mathrm{2}\theta\right)+{cos}\left(\mathrm{4}\theta\right)\right\}{d}\theta \\ $$$$=\frac{\mathrm{3}}{\mathrm{8}}\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{2}\theta\right){d}\theta\:+\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:{cos}\left(\mathrm{4}\theta\right){d}\theta \\ $$$$=\frac{\mathrm{3}\pi}{\mathrm{32}}\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sin}\left(\mathrm{2}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:+\frac{\mathrm{1}}{\mathrm{32}}\left[{sin}\left(\mathrm{4}\theta\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \\ $$$$=\frac{\mathrm{3}\pi}{\mathrm{32}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:+\mathrm{0}\:\Rightarrow\:{I}\:=\frac{\mathrm{3}\pi}{\mathrm{32}}\:+\frac{\mathrm{1}}{\mathrm{4}} \\ $$