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BeMath-1-find-1-2-2-0-pi-2-x-sin-x-1-cos-x-2-dx-




Question Number 108450 by bemath last updated on 17/Aug/20
  ((BeMath)/(⊂⊃))  (1)find ((1/2))!  (2)∫_0 ^(π/2) ((x sin x)/((1+cos x)^2 )) dx
$$\:\:\frac{\mathcal{B}{e}\mathcal{M}{ath}}{\subset\supset} \\ $$$$\left(\mathrm{1}\right){find}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)! \\ $$$$\left(\mathrm{2}\right)\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\frac{{x}\:\mathrm{sin}\:{x}}{\left(\mathrm{1}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\:{dx} \\ $$
Commented by Smail last updated on 17/Aug/20
I_n =∫_0 ^∞ t^n e^(−t) dt=n!  n=(1/2)  I_(1/2) =∫_0 ^∞ (√t)e^(−t) dt=((1/2))!  x=(√t)⇒dx=(dt/(2(√t)))  I_(1/2) =2∫_0 ^∞ x^2 e^(−x^2 ) dx=((1/2))!  By parts  u=x⇒u′=1  v′=xe^(−x^2 ) ⇒v=((−1)/2)e^(−x^2 )   I_(1/2) =∫_0 ^∞ e^(−x^2 ) dx=((1/2))!  ∫_0 ^∞ e^(−x^2 ) dx=((√π)/2)  So  ((1/2))!=((√π)/2)
$${I}_{{n}} =\int_{\mathrm{0}} ^{\infty} {t}^{{n}} {e}^{−{t}} {dt}={n}! \\ $$$${n}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${I}_{\mathrm{1}/\mathrm{2}} =\int_{\mathrm{0}} ^{\infty} \sqrt{{t}}{e}^{−{t}} {dt}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)! \\ $$$${x}=\sqrt{{t}}\Rightarrow{dx}=\frac{{dt}}{\mathrm{2}\sqrt{{t}}} \\ $$$${I}_{\mathrm{1}/\mathrm{2}} =\mathrm{2}\int_{\mathrm{0}} ^{\infty} {x}^{\mathrm{2}} {e}^{−{x}^{\mathrm{2}} } {dx}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)! \\ $$$${By}\:{parts} \\ $$$${u}={x}\Rightarrow{u}'=\mathrm{1} \\ $$$${v}'={xe}^{−{x}^{\mathrm{2}} } \Rightarrow{v}=\frac{−\mathrm{1}}{\mathrm{2}}{e}^{−{x}^{\mathrm{2}} } \\ $$$${I}_{\mathrm{1}/\mathrm{2}} =\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\left(\frac{\mathrm{1}}{\mathrm{2}}\right)! \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{x}^{\mathrm{2}} } {dx}=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$$${So} \\ $$$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)!=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$
Commented by bemath last updated on 17/Aug/20
thank you
$${thank}\:{you} \\ $$
Answered by john santu last updated on 17/Aug/20
Answered by Dwaipayan Shikari last updated on 17/Aug/20
((1/2))!=(1/2)Γ((1/2))=((√π)/2)
$$\left(\frac{\mathrm{1}}{\mathrm{2}}\right)!=\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}} \\ $$

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