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Question-136651




Question Number 136651 by mhabs last updated on 24/Mar/21
Answered by Ñï= last updated on 24/Mar/21
∫_0 ^(π/2) (1/(1+(atan x)^2 ))dx=^(t=tan x) ∫_0 ^∞ (1/(1+a^2 t^2 ))∙(1/(1+t^2 ))dt  =(1/(1−a^2 ))∫_0 ^∞ (((−a^2 )/(1+a^2 t^2 ))+(1/(1+t^2 )))dt=(1/(1−a^2 )){(π/2)−(a^2 /a)tan^(−1) (at)∣_0 ^∞ }  =(1/(1−a^2 ))(1−a)(π/2)  =(π/(2(1+a)))
$$\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{1}}{\mathrm{1}+\left({a}\mathrm{tan}\:{x}\right)^{\mathrm{2}} }{dx}\overset{{t}=\mathrm{tan}\:{x}} {=}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{a}^{\mathrm{2}} {t}^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \left(\frac{−{a}^{\mathrm{2}} }{\mathrm{1}+{a}^{\mathrm{2}} {t}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\right){dt}=\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }\left\{\frac{\pi}{\mathrm{2}}−\frac{{a}^{\mathrm{2}} }{{a}}\mathrm{tan}^{−\mathrm{1}} \left({at}\right)\mid_{\mathrm{0}} ^{\infty} \right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{a}^{\mathrm{2}} }\left(\mathrm{1}−{a}\right)\frac{\pi}{\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{2}\left(\mathrm{1}+{a}\right)} \\ $$
Answered by Dwaipayan Shikari last updated on 24/Mar/21
∫_0 ^(π/2) (1/(1+(atan(x))^2 ))dx      atan(x)=u  asec^2 (x)=(du/dx)  =a∫_0 ^∞ (1/((1+u^2 )(a^2 +u^2 )))du=(a/(a^2 −1))∫_0 ^∞ (1/(1+u^2 ))−(1/(a^2 +u^2 ))du  =((πa)/(2(a^2 −1)))−(π/2)=(π/(2(a+1)))
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}}{\mathrm{1}+\left({atan}\left({x}\right)\right)^{\mathrm{2}} }{dx}\:\:\:\:\:\:{atan}\left({x}\right)={u}\:\:{asec}^{\mathrm{2}} \left({x}\right)=\frac{{du}}{{dx}} \\ $$$$={a}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} +{u}^{\mathrm{2}} \right)}{du}=\frac{{a}}{{a}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }−\frac{\mathrm{1}}{{a}^{\mathrm{2}} +{u}^{\mathrm{2}} }{du} \\ $$$$=\frac{\pi{a}}{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}−\frac{\pi}{\mathrm{2}}=\frac{\pi}{\mathrm{2}\left({a}+\mathrm{1}\right)} \\ $$

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