Question Number 121602 by benjo_mathlover last updated on 09/Nov/20
Answered by liberty last updated on 10/Nov/20
![I=∫_0 ^(π/2) (dx/(b^2 tan^2 x+1)) let φ=b tan x ⇒dφ=bsec^2 x dx dx = (φ/(φ^2 +1)) dφ I=∫_0 ^∞ ((φ dφ)/((b^2 +φ^2 )(1+φ^2 ))) = (b/(b^2 −1))∫_0 ^∞ [ (1/(1+φ^2 ))−(1/(b^2 +φ^2 )) ] dφ I= (b/(b^2 −1)) [ tan^(−1) (φ)−tan^(−1) ((φ/b)) ]_0 ^∞ I= (b/(b^2 −1)) [ (π/2) − (π/(2b)) ] I=(π/2) (1−(1/b))((b/(b^2 −1))) = (π/2)(((b−1)/b))((b/((b−1)(b+1)))) I= (π/(2(b+1))) .▲](https://www.tinkutara.com/question/Q121606.png)
$$\mathrm{I}=\underset{\mathrm{0}} {\overset{\pi/\mathrm{2}} {\int}}\:\frac{\mathrm{dx}}{\mathrm{b}^{\mathrm{2}} \:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}+\mathrm{1}} \\ $$$$\mathrm{let}\:\phi=\mathrm{b}\:\mathrm{tan}\:\mathrm{x}\:\Rightarrow\mathrm{d}\phi=\mathrm{bsec}\:^{\mathrm{2}} \mathrm{x}\:\mathrm{dx}\: \\ $$$$\:\mathrm{dx}\:=\:\frac{\phi}{\phi^{\mathrm{2}} +\mathrm{1}}\:\mathrm{d}\phi\: \\ $$$$\mathrm{I}=\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\phi\:\mathrm{d}\phi}{\left(\mathrm{b}^{\mathrm{2}} +\phi^{\mathrm{2}} \right)\left(\mathrm{1}+\phi^{\mathrm{2}} \right)}\:=\:\frac{\mathrm{b}}{\mathrm{b}^{\mathrm{2}} −\mathrm{1}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\left[\:\frac{\mathrm{1}}{\mathrm{1}+\phi^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{b}^{\mathrm{2}} +\phi^{\mathrm{2}} }\:\right]\:\mathrm{d}\phi \\ $$$$\mathrm{I}=\:\frac{\mathrm{b}}{\mathrm{b}^{\mathrm{2}} −\mathrm{1}}\:\left[\:\mathrm{tan}^{−\mathrm{1}} \left(\phi\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\phi}{\mathrm{b}}\right)\:\right]_{\mathrm{0}} ^{\infty} \\ $$$$\mathrm{I}=\:\frac{\mathrm{b}}{\mathrm{b}^{\mathrm{2}} −\mathrm{1}}\:\left[\:\frac{\pi}{\mathrm{2}}\:−\:\frac{\pi}{\mathrm{2b}}\:\right] \\ $$$$\mathrm{I}=\frac{\pi}{\mathrm{2}}\:\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{b}}\right)\left(\frac{\mathrm{b}}{\mathrm{b}^{\mathrm{2}} −\mathrm{1}}\right)\:=\:\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{b}−\mathrm{1}}{\mathrm{b}}\right)\left(\frac{\mathrm{b}}{\left(\mathrm{b}−\mathrm{1}\right)\left(\mathrm{b}+\mathrm{1}\right)}\right) \\ $$$$\mathrm{I}=\:\frac{\pi}{\mathrm{2}\left(\mathrm{b}+\mathrm{1}\right)}\:.\blacktriangle \\ $$
Answered by Olaf last updated on 10/Nov/20
![1) b = 0 I = ∫_0 ^(π/2) dx = (π/2) 2) b = ±1 I = ∫_0 ^(π/2) (dx/(1+tan^2 x)) I = ∫_0 ^(π/2) cos^2 xdx I = ∫_0 ^(π/2) ((1+cos(2x))/2)dx I = [(1/2)x+(1/4)sin(2x)]_0 ^(π/2) I = (π/4) 3) General case : b ≠ 0, b ≠ ±1 u = tanx du = (1+tan^2 x)dx = (1+u^2 )dx I = ∫_0 ^∞ (du/((b^2 u^2 +1)(u^2 +1))) I = (1/(b^2 −1))∫_0 ^∞ [(b^2 /(b^2 u^2 +1))−(1/(1+u^2 ))]du I = (1/(b^2 −1))∫_0 ^∞ [(1/((1/b^2 )+u^2 ))−(1/(1+u^2 ))]du I = (1/(b^2 −1))[barctan(bu)−arctanu]_0 ^∞ I = (1/(b^2 −1))[(b×sign(b)(π/2)−(π/2))−(b(π/4)−(π/4))] I = (1/(b^2 −1))[(π/2)(∣b∣−1)−(π/4)(b−1)] If b>0 : I = (1/(b^2 −1))[(π/2)(b−1)−(π/4)(b−1)] I = (1/(b+1))[(π/2)−(π/4)] = (π/(4(b+1))) If b<0 : I = (1/(b^2 −1))[(π/2)(−b−1)−(π/4)(b−1)] I = (π/(4(b^2 −1)))[2(−b−1)−(b−1)] I = (π/(4(b^2 −1)))(−3b−3) I = −((3π)/(4(b−1)))](https://www.tinkutara.com/question/Q121607.png)
$$\left.\mathrm{1}\right)\:{b}\:=\:\mathrm{0} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}\:=\:\frac{\pi}{\mathrm{2}} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:{b}\:=\:\pm\mathrm{1} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{dx}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \mathrm{cos}^{\mathrm{2}} {xdx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+\mathrm{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}{dx} \\ $$$$\mathrm{I}\:=\:\left[\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2}{x}\right)\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\ $$$$\mathrm{I}\:=\:\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$$\left.\mathrm{3}\right)\:\mathrm{General}\:\mathrm{case}\::\:{b}\:\neq\:\mathrm{0},\:{b}\:\neq\:\pm\mathrm{1} \\ $$$${u}\:=\:\mathrm{tan}{x} \\ $$$${du}\:=\:\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}\right){dx}\:=\:\left(\mathrm{1}+{u}^{\mathrm{2}} \right){dx} \\ $$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\infty} \frac{{du}}{\left({b}^{\mathrm{2}} {u}^{\mathrm{2}} +\mathrm{1}\right)\left({u}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left[\frac{{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} {u}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\right]{du} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \left[\frac{\mathrm{1}}{\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+{u}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\right]{du} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} −\mathrm{1}}\left[{b}\mathrm{arctan}\left({bu}\right)−\mathrm{arctan}{u}\right]_{\mathrm{0}} ^{\infty} \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} −\mathrm{1}}\left[\left({b}×\mathrm{sign}\left({b}\right)\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{2}}\right)−\left({b}\frac{\pi}{\mathrm{4}}−\frac{\pi}{\mathrm{4}}\right)\right] \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} −\mathrm{1}}\left[\frac{\pi}{\mathrm{2}}\left(\mid{b}\mid−\mathrm{1}\right)−\frac{\pi}{\mathrm{4}}\left({b}−\mathrm{1}\right)\right] \\ $$$$\mathrm{If}\:{b}>\mathrm{0}\:: \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} −\mathrm{1}}\left[\frac{\pi}{\mathrm{2}}\left({b}−\mathrm{1}\right)−\frac{\pi}{\mathrm{4}}\left({b}−\mathrm{1}\right)\right] \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{{b}+\mathrm{1}}\left[\frac{\pi}{\mathrm{2}}−\frac{\pi}{\mathrm{4}}\right]\:=\:\frac{\pi}{\mathrm{4}\left({b}+\mathrm{1}\right)} \\ $$$$\mathrm{If}\:{b}<\mathrm{0}\:: \\ $$$$\mathrm{I}\:=\:\frac{\mathrm{1}}{{b}^{\mathrm{2}} −\mathrm{1}}\left[\frac{\pi}{\mathrm{2}}\left(−{b}−\mathrm{1}\right)−\frac{\pi}{\mathrm{4}}\left({b}−\mathrm{1}\right)\right] \\ $$$$\mathrm{I}\:=\:\frac{\pi}{\mathrm{4}\left({b}^{\mathrm{2}} −\mathrm{1}\right)}\left[\mathrm{2}\left(−{b}−\mathrm{1}\right)−\left({b}−\mathrm{1}\right)\right] \\ $$$$\mathrm{I}\:=\:\frac{\pi}{\mathrm{4}\left({b}^{\mathrm{2}} −\mathrm{1}\right)}\left(−\mathrm{3}{b}−\mathrm{3}\right) \\ $$$$\mathrm{I}\:=\:−\frac{\mathrm{3}\pi}{\mathrm{4}\left({b}−\mathrm{1}\right)} \\ $$