Question Number 207565 by mathzup last updated on 18/May/24
$${find}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{x}^{\mathrm{2}} }{{tan}^{\mathrm{2}} {x}}{dx} \\ $$
Answered by sniper237 last updated on 19/May/24
![∫_0 ^(π/2) ((x^2 (1−sin^2 x))/(sin^2 x)) dx=∫_0 ^(π/2) (x^2 /(sin^2 x))dx−(π^3 /(24)) =^(PI) [−(x^2 /(tanx))]_(→0) +2∫_0 ^(π/2) (x/(tanx))dx−(π^3 /(24)) =^(PI) 2[xln(sinx)]_(→0) −2∫_0 ^(π/2) ln(sinx)dx−(π^3 /(24)) =^(t=sinx) −2∫_0 ^1 ((lnt)/( (√(1−t^2 ))))dt −(π^3 /(24)) =−2∂_a (∫_0 ^1 (t^a /( (√(1−t^2 ))))dt)_0 −(π^3 /(24)) =^(u=t^2 ) −∂_a (∫_0 ^1 u^((a−1)/2) (1−u)^((1/2)−1) du)−(π^3 /(24)) =−∂_a (((Γ(((a+1)/2))Γ((1/2)))/(Γ((a/2)+1))))_0 −(π^3 /(24)) = −(1/2) ((Γ′(1/2)Γ(1/2)−Γ′(1)Γ(1/2)^2 )/1^2 )−(π^3 /(24)) =((3πγ−6(√π) Γ′(1/2)−π^3 )/(24))](https://www.tinkutara.com/question/Q207580.png)
$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{x}^{\mathrm{2}} \left(\mathrm{1}−{sin}^{\mathrm{2}} {x}\right)}{{sin}^{\mathrm{2}} {x}}\:{dx}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{x}^{\mathrm{2}} }{{sin}^{\mathrm{2}} {x}}{dx}−\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$$$\overset{{PI}} {=}\left[−\frac{{x}^{\mathrm{2}} }{{tanx}}\right]_{\rightarrow\mathrm{0}} +\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{x}}{{tanx}}{dx}−\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$$$\overset{{PI}} {=}\:\mathrm{2}\left[{xln}\left({sinx}\right)\right]_{\rightarrow\mathrm{0}} −\mathrm{2}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {ln}\left({sinx}\right){dx}−\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$$$\overset{{t}={sinx}} {=}−\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\:−\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$$$=−\mathrm{2}\partial_{{a}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{t}^{{a}} }{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{dt}\right)_{\mathrm{0}} −\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$$$\overset{{u}={t}^{\mathrm{2}} } {=}−\partial_{{a}} \left(\int_{\mathrm{0}} ^{\mathrm{1}} {u}^{\frac{{a}−\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{u}\right)^{\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{1}} {du}\right)−\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$$$=−\partial_{{a}} \left(\frac{\Gamma\left(\frac{{a}+\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\frac{{a}}{\mathrm{2}}+\mathrm{1}\right)}\right)_{\mathrm{0}} −\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\Gamma'\left(\mathrm{1}/\mathrm{2}\right)\Gamma\left(\mathrm{1}/\mathrm{2}\right)−\Gamma'\left(\mathrm{1}\right)\Gamma\left(\mathrm{1}/\mathrm{2}\right)^{\mathrm{2}} }{\mathrm{1}^{\mathrm{2}} }−\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$$$=\frac{\mathrm{3}\pi\gamma−\mathrm{6}\sqrt{\pi}\:\Gamma'\left(\mathrm{1}/\mathrm{2}\right)−\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$
Answered by Berbere last updated on 19/May/24
![x→(π/2)−x =∫_0 ^(π/2) ((π/2)−x)^2 tan^2 (x)dx=∫_0 ^(π/2) (1+tan^2 (x))((π/2)−x)^2 dx −∫_0 ^(π/2) ((π/2)−x)^2 =A−B B=−(1/3)[((π/2)−x)^3 ]_0 ^(π/2) =(π^3 /(24)) A=[tan (x)((π/2)−x)^2 ]_0 ^(π/2) −2∫_0 ^(π/2) tan (x)((π/2)−x) =−2∫_0 ^(π/2) tan (x)((π/2)−x)dx=[2ln(cos(x))((π/2)−x)]_0 ^(π/2) −2∫_0 ^(π/2) ln(cos(x))dx=−2.−(π/2)ln(2)=πln(2) πln(2)−(π^3 /(24))](https://www.tinkutara.com/question/Q207581.png)
$${x}\rightarrow\frac{\pi}{\mathrm{2}}−{x} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \left({x}\right){dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \left({x}\right)\right)\left(\frac{\pi}{\mathrm{2}}−{x}\right)^{\mathrm{2}} {dx} \\ $$$$−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\frac{\pi}{\mathrm{2}}−{x}\right)^{\mathrm{2}} ={A}−{B} \\ $$$${B}=−\frac{\mathrm{1}}{\mathrm{3}}\left[\left(\frac{\pi}{\mathrm{2}}−{x}\right)^{\mathrm{3}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$$${A}=\left[\mathrm{tan}\:\left({x}\right)\left(\frac{\pi}{\mathrm{2}}−{x}\right)^{\mathrm{2}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}\:\left({x}\right)\left(\frac{\pi}{\mathrm{2}}−{x}\right) \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{tan}\:\left({x}\right)\left(\frac{\pi}{\mathrm{2}}−{x}\right){dx}=\left[\mathrm{2}{ln}\left({cos}\left({x}\right)\right)\left(\frac{\pi}{\mathrm{2}}−{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$−\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cos}\left({x}\right)\right){dx}=−\mathrm{2}.−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)=\pi{ln}\left(\mathrm{2}\right) \\ $$$$\pi{ln}\left(\mathrm{2}\right)−\frac{\pi^{\mathrm{3}} }{\mathrm{24}} \\ $$