Question Number 32040 by abdo imad last updated on 18/Mar/18
$${let}\:{give}\:{f}\left({x}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+{x}\:{tant}}\:\: \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{tant}}{\left(\mathrm{1}+{xtant}\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right){give}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{tant}}{\left(\mathrm{1}+\sqrt{\mathrm{3}}\:{tant}\right)^{\mathrm{2}} }\:{dt}\:. \\ $$
Commented by abdo imad last updated on 21/Mar/18
![ch.tant =u give f(x)=∫_0 ^∞ (1/(1+xu^2 )) (du/(1+u^2 )) =∫_0 ^∞ (du/((1+u^2 )(1+xu^2 ))) let decompose F(u)= (1/((1+u^2 )(1+xu^2 ))) = ((au +b)/(1+u^2 )) + ((cu +d)/(1+xu^2 )) F(−u)=F(u)⇒a=c=0⇒F(u)= (b/(1+u^2 )) +(d/(1+xu^2 )) F(0)=1= b+d F(1)= (1/(2(1+x))) = (b/2) + (d/(1+x))= (b/2) +((2d)/(2(1+x))) ⇒ (1/(1+x)) = b +((2d)/(1+x)) ⇒1 =(1+x)b +2d ⇒ 1 =(1+x)b +2(1−b)⇒1= 2 +(x−1)b ⇒b=(1/(1−x)) d=1−b =1−(1/(1−x)) = ((−x)/(1−x)) ⇒ F(u) = (1/((1−x)(1+u^2 ))) −(x/((1−x)(1+xu^2 ))) f(x)= (1/(1−x))∫_0 ^∞ (du/(1+u^2 )) −(x/(1−x))∫_0 ^∞ (du/(1+xu^2 )) but we have ∫_0 ^∞ (du/(1+u^2 ))= [arctanu]_0 ^∞ =(π/2) and ch. (√x) u =t (we suppose x>0) ∫_0 ^∞ (du/(1+xu^2 )) = ∫_0 ^∞ (1/(1+t^2 )) (dt/( (√x))) = (π/(2(√x))) ⇒ f(x)= (π/(2(1−x))) − ((πx)/(2(√x)(1−x))) =(π/(2(1−x)))(1− (√x)) =(π/2) ((1−(√x))/((1+(√x))(1−(√x)))) = (π/(2(1+(√x)))) . 2)we have f(x)=∫_0 ^(π/2) (dt/(1+xtant)) ⇒ f^′ (x) =−∫_0 ^(π/2) ((tant)/((1+xtant)^2 ))dt ⇒ ∫_0 ^(π/2) ((tant)/((1+xtant)^2 )) dt =−f^′ (x) but f^′ (x) =−(π/2) (((1+(√(x )))^′ )/((1+(√x))^2 )) =−(π/2) (1/(2(√x)(1+(√x))^2 )) ⇒ ∫_0 ^(π/2) ((tant)/((1+xtant)^2 )) dt = (π/(4 (√x)(1+(√x))^2 )) 3) let take x=(√3) weget ∫_0 ^(π/2) ((tant)/((1+(√3)tant)^2 )) dt = (π/(4^4 (√3)(1+^4 (√3))^2 )) .](https://www.tinkutara.com/question/Q32198.png)
$${ch}.{tant}\:={u}\:{give}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{xu}^{\mathrm{2}} }\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{xu}^{\mathrm{2}} \right)}\:\:{let}\:{decompose} \\ $$$${F}\left({u}\right)=\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\mathrm{1}+{xu}^{\mathrm{2}} \right)}\:=\:\:\frac{{au}\:+{b}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\:\frac{{cu}\:+{d}}{\mathrm{1}+{xu}^{\mathrm{2}} } \\ $$$${F}\left(−{u}\right)={F}\left({u}\right)\Rightarrow{a}={c}=\mathrm{0}\Rightarrow{F}\left({u}\right)=\:\frac{{b}}{\mathrm{1}+{u}^{\mathrm{2}} }\:+\frac{{d}}{\mathrm{1}+{xu}^{\mathrm{2}} } \\ $$$${F}\left(\mathrm{0}\right)=\mathrm{1}=\:{b}+{d} \\ $$$${F}\left(\mathrm{1}\right)=\:\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{x}\right)}\:=\:\frac{{b}}{\mathrm{2}}\:+\:\frac{{d}}{\mathrm{1}+{x}}=\:\frac{{b}}{\mathrm{2}}\:+\frac{\mathrm{2}{d}}{\mathrm{2}\left(\mathrm{1}+{x}\right)}\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}}\:=\:{b}\:+\frac{\mathrm{2}{d}}{\mathrm{1}+{x}}\:\Rightarrow\mathrm{1}\:=\left(\mathrm{1}+{x}\right){b}\:+\mathrm{2}{d}\:\:\Rightarrow \\ $$$$\mathrm{1}\:=\left(\mathrm{1}+{x}\right){b}\:+\mathrm{2}\left(\mathrm{1}−{b}\right)\Rightarrow\mathrm{1}=\:\mathrm{2}\:\:+\left({x}−\mathrm{1}\right){b}\:\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{1}−{x}} \\ $$$${d}=\mathrm{1}−{b}\:=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}−{x}}\:=\:\frac{−{x}}{\mathrm{1}−{x}}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{u}^{\mathrm{2}} \right)}\:−\frac{{x}}{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{xu}^{\mathrm{2}} \right)} \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}−{x}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:−\frac{{x}}{\mathrm{1}−{x}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{du}}{\mathrm{1}+{xu}^{\mathrm{2}} }\:{but}\:{we}\:{have} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }=\:\left[{arctanu}\right]_{\mathrm{0}} ^{\infty} \:=\frac{\pi}{\mathrm{2}}\:{and}\:{ch}.\:\sqrt{{x}}\:{u}\:={t}\:\:\left({we}\:{suppose}\:{x}>\mathrm{0}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{xu}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\:\frac{{dt}}{\:\sqrt{{x}}}\:=\:\frac{\pi}{\mathrm{2}\sqrt{{x}}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\:\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{x}\right)}\:−\:\:\frac{\pi{x}}{\mathrm{2}\sqrt{{x}}\left(\mathrm{1}−{x}\right)}\:=\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{x}\right)}\left(\mathrm{1}−\:\sqrt{{x}}\right) \\ $$$$=\frac{\pi}{\mathrm{2}}\:\frac{\mathrm{1}−\sqrt{{x}}}{\left(\mathrm{1}+\sqrt{{x}}\right)\left(\mathrm{1}−\sqrt{{x}}\right)}\:=\:\frac{\pi}{\mathrm{2}\left(\mathrm{1}+\sqrt{{x}}\right)}\:. \\ $$$$\left.\mathrm{2}\right){we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{dt}}{\mathrm{1}+{xtant}}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=−\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{tant}}{\left(\mathrm{1}+{xtant}\right)^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{tant}}{\left(\mathrm{1}+{xtant}\right)^{\mathrm{2}} }\:{dt}\:=−{f}^{'} \left({x}\right)\:{but} \\ $$$${f}^{'} \left({x}\right)\:=−\frac{\pi}{\mathrm{2}}\:\frac{\left(\mathrm{1}+\sqrt{{x}\:}\right)^{'} }{\left(\mathrm{1}+\sqrt{{x}}\right)^{\mathrm{2}} }\:=−\frac{\pi}{\mathrm{2}}\:\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}}\left(\mathrm{1}+\sqrt{{x}}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{tant}}{\left(\mathrm{1}+{xtant}\right)^{\mathrm{2}} }\:{dt}\:=\:\:\:\frac{\pi}{\mathrm{4}\:\sqrt{{x}}\left(\mathrm{1}+\sqrt{{x}}\right)^{\mathrm{2}} } \\ $$$$\left.\mathrm{3}\right)\:{let}\:{take}\:{x}=\sqrt{\mathrm{3}}\:\:{weget} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{tant}}{\left(\mathrm{1}+\sqrt{\mathrm{3}}{tant}\right)^{\mathrm{2}} }\:{dt}\:=\:\:\frac{\pi}{\mathrm{4}^{\mathrm{4}} \sqrt{\mathrm{3}}\left(\mathrm{1}+\:^{\mathrm{4}} \sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\:. \\ $$
Commented by abdo imad last updated on 21/Mar/18
$${x}>\mathrm{0}. \\ $$