Question Number 32354 by abdo imad last updated on 23/Mar/18
$${calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{sin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }\:. \\ $$
Commented by abdo imad last updated on 28/Mar/18
![let put I = ∫_0 ^(π/4) (dt/((1+sin^2 t)^2 )) I = ∫_0 ^(π/4) (dt/((1+((1−cos(2t))/2))^2 )) = 4∫_0 ^(π/4) (dt/((3−cos(2t))^2 )) =_(2t=x) 4 ∫_0 ^(π/2) (1/((3−cosx)^2 )) (dx/2) = 2 ∫_0 ^(π/2) (dx/((3−cosx)^2 )) ch. tan((x/2)) =tgive I = 2 ∫_0 ^1 (1/((3 −((1−t^2 )/(1+t^2 )))^2 )) ((2dt)/(1+t^2 )) = 4 ∫_0 ^1 (dt/((1+t^2 ) (((3 +3t^2 −1+t^2 )^2 )/((1+t^2 )^2 )))) = 4 ∫_0 ^1 ((1+t^2 )/((2+4t^2 )^2 ))dt = ∫_0 ^1 ((1+t^2 )/((1+2t^2 )^2 )) dt = ∫_0 ^1 ((1+2t^2 −t^2 )/((1+2t^2 )^2 ))dt = ∫_0 ^1 (dt/((1+2t^2 ))) dt − ∫_0 ^1 (t^2 /((1+2t^2 )^2 ))dt ∫_0 ^1 (dt/(1+2t^2 ))dt =_(t(√2) =u) ∫_0 ^(√2) (1/(1+u^2 )) (du/( (√2))) = (1/( (√2))) artan((√2)) ∫_0 ^1 (t^2 /((1+2t^2 )^2 ))dt =_(t(√2)=u) ∫_0 ^(√2) (1/((1+u^2 )^2 )) (u^2 /2) (du/( (√2))) = (1/(2(√2))) ∫_0 ^(√2) (u^2 /((1+u^2 )^2 ))du but ∫_0 ^(√2) (u^2 /((1+u^2 )^2 ))du =−(1/2) ∫_0 ^(√2) u.(((−2u)/((1+u^2 )^2 )))du by parts =−(1/2) ( [ (u/(1+u^2 ))]_0 ^(√2) −∫_0 ^(√2) (u/(1+u^2 )) du) =−(1/2) ((√2)/3) +(1/4) ∫_0 ^(√2) ((2u)/(1+u^2 ))du =((−(√2))/6) +(1/4) [ln(1+u^2 )]_0 ^(√2) = ((−(√2))/6) +(1/4)ln(3) ⇒∫_0 ^1 (t^2 /((1+2t^2 )^2 ))dt =(1/(2(√2)))(((−(√2))/6) +(1/4)ln3) = ((−1)/(12)) +((ln3)/(8(√2))) I = (1/( (√2))) arctan((√2)) +(1/(12)) −((ln3)/(8(√2))) .](https://www.tinkutara.com/question/Q32587.png)
$${let}\:{put}\:{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dt}}{\left(\mathrm{1}+{sin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} } \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{dt}}{\left(\mathrm{1}+\frac{\mathrm{1}−{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right)^{\mathrm{2}} }\:=\:\mathrm{4}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\frac{{dt}}{\left(\mathrm{3}−{cos}\left(\mathrm{2}{t}\right)\right)^{\mathrm{2}} } \\ $$$$=_{\mathrm{2}{t}={x}} \:\mathrm{4}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}−{cosx}\right)^{\mathrm{2}} }\:\frac{{dx}}{\mathrm{2}}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dx}}{\left(\mathrm{3}−{cosx}\right)^{\mathrm{2}} } \\ $$$${ch}.\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:={tgive} \\ $$$${I}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{3}\:−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\:\mathrm{4}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:\frac{\left(\mathrm{3}\:+\mathrm{3}{t}^{\mathrm{2}} −\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }} \\ $$$$=\:\mathrm{4}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\left(\mathrm{2}+\mathrm{4}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)}\:{dt}\:\:−\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} }{dt}\:=_{{t}\sqrt{\mathrm{2}}\:={u}} \:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\:\frac{{du}}{\:\sqrt{\mathrm{2}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{artan}\left(\sqrt{\mathrm{2}}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=_{{t}\sqrt{\mathrm{2}}={u}} \:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\:\frac{{u}^{\mathrm{2}} }{\mathrm{2}}\:\frac{{du}}{\:\sqrt{\mathrm{2}}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\:\frac{{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{du}\:\:{but}\:\:\:\:\:\: \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{u}^{\mathrm{2}} }{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }{du}\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:{u}.\left(\frac{−\mathrm{2}{u}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)^{\mathrm{2}} }\right){du}\:{by}\:{parts} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\left(\:\left[\:\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:−\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{u}}{\mathrm{1}+{u}^{\mathrm{2}} }\:{du}\right) \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\:+\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \:\:\frac{\mathrm{2}{u}}{\mathrm{1}+{u}^{\mathrm{2}} }{du}\:=\frac{−\sqrt{\mathrm{2}}}{\mathrm{6}}\:\:+\frac{\mathrm{1}}{\mathrm{4}}\:\left[{ln}\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{2}}} \\ $$$$=\:\frac{−\sqrt{\mathrm{2}}}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{3}\right)\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{\mathrm{2}} }{\left(\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\left(\frac{−\sqrt{\mathrm{2}}}{\mathrm{6}}\:+\frac{\mathrm{1}}{\mathrm{4}}{ln}\mathrm{3}\right) \\ $$$$=\:\frac{−\mathrm{1}}{\mathrm{12}}\:+\frac{{ln}\mathrm{3}}{\mathrm{8}\sqrt{\mathrm{2}}} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{arctan}\left(\sqrt{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{12}}\:−\frac{{ln}\mathrm{3}}{\mathrm{8}\sqrt{\mathrm{2}}}\:. \\ $$