Question Number 45213 by MrW3 last updated on 10/Oct/18
$${Find}\:\int\sqrt{{a}^{\mathrm{2}} −\mathrm{tan}^{\mathrm{2}} \:{x}}\:{dx}\:\left({with}\:{a}>\mathrm{0}\right) \\ $$$$\left({related}\:{to}\:{Q}\mathrm{45187}\right) \\ $$
Answered by MrW3 last updated on 10/Oct/18
![I=∫(√(a^2 −tan^2 x)) dx let u=tan x du=(dx/(cos^2 x))=(1+tan^2 x)dx=(1+u^2 )dx I=∫((√(a^2 −u^2 ))/(1+u^2 ))du let u=a sin θ ⇒θ=sin^(−1) (u/a) du=a cos θ dθ I=a^2 ∫((cos^2 θ)/(1+a^2 sin^2 θ))dθ let v=tan θ dv=(dθ/(cos^2 θ))=(1+tan^2 θ)dθ=(v^2 +1)dθ I=a^2 ∫(dv/({(a^2 +1)v^2 +1}(v^2 +1))) (1/({(a^2 +1)v^2 +1}(v^2 +1)))=(A/((a^2 +1)v^2 +1))+(B/(v^2 +1)) A(v^2 +1)+B(a^2 +1)v^2 +B=1 {A+B(a^2 +1)}v^2 +(A+B)=1 ⇒A+B=1 ⇒A+B(a^2 +1)=0 ⇒B=−(1/a^2 ) ⇒A=((a^2 +1)/a^2 ) I=a^2 ∫[((a^2 +1)/(a^2 {(a^2 +1)v^2 +1}))−(1/(a^2 (v^2 +1)))]dv I=∫[((a^2 +1)/((a^2 +1)v^2 +1))−(1/(v^2 +1))]dv I=(a^2 +1)∫(1/((a^2 +1)v^2 +1))dv−∫(1/(v^2 +1))dv I=(a^2 +1)I_1 −I_2 I_1 =∫(1/((a^2 +1)v^2 +1))dv let p=(√(a^2 +1))v dp=(√(a^2 +1)) dv I_1 =(1/( (√(a^2 +1))))∫(dp/(p^2 +1))=(1/( (√(a^2 +1))))tan^(−1) p I_1 =(1/( (√(a^2 +1)))) tan^(−1) ((√(a^2 +1))v) I_1 =(1/( (√(a^2 +1)))) tan^(−1) ((√(a^2 +1)) tan θ) I_1 =(1/( (√(a^2 +1)))) tan^(−1) ((√(a^2 +1)) ((sin θ)/( (√(1−sin^2 θ))))) I_1 =(1/( (√(a^2 +1)))) tan^(−1) ((√(a^2 +1)) ((u/a)/( (√(1−(u^2 /a^2 )))))) I_1 =(1/( (√(a^2 +1)))) tan^(−1) ((√(a^2 +1)) (u/( (√(a^2 −u^2 ))))) ⇒I_1 =(1/( (√(a^2 +1)))) tan^(−1) ((((√(a^2 +1)) tan x)/( (√(a^2 −tan^2 x))))) I_2 =∫(1/(v^2 +1))dv=tan^(−1) v=θ=sin^(−1) (u/a) ⇒I_2 =sin^(−1) (u/a)=sin^(−1) (((tan x)/a)) I=(a^2 +1)I_1 −I_2 ⇒I=((√(a^2 +1))) tan^(−1) ((((√(a^2 +1)) tan x)/( (√(a^2 −tan^2 x)))))−sin^(−1) (((tan x)/a))+C](https://www.tinkutara.com/question/Q45217.png)
$${I}=\int\sqrt{{a}^{\mathrm{2}} −\mathrm{tan}^{\mathrm{2}} \:{x}}\:{dx} \\ $$$${let}\:{u}=\mathrm{tan}\:{x} \\ $$$${du}=\frac{{dx}}{\mathrm{cos}^{\mathrm{2}} \:{x}}=\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:{x}\right){dx}=\left(\mathrm{1}+{u}^{\mathrm{2}} \right){dx} \\ $$$${I}=\int\frac{\sqrt{{a}^{\mathrm{2}} −{u}^{\mathrm{2}} }}{\mathrm{1}+{u}^{\mathrm{2}} }{du} \\ $$$${let}\:{u}={a}\:\mathrm{sin}\:\theta\:\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \frac{{u}}{{a}} \\ $$$${du}={a}\:\mathrm{cos}\:\theta\:{d}\theta \\ $$$${I}={a}^{\mathrm{2}} \int\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\mathrm{1}+{a}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta}{d}\theta \\ $$$${let}\:{v}=\mathrm{tan}\:\theta \\ $$$${dv}=\frac{{d}\theta}{\mathrm{cos}^{\mathrm{2}} \:\theta}=\left(\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\theta\right){d}\theta=\left({v}^{\mathrm{2}} +\mathrm{1}\right){d}\theta \\ $$$${I}={a}^{\mathrm{2}} \int\frac{{dv}}{\left\{\left({a}^{\mathrm{2}} +\mathrm{1}\right){v}^{\mathrm{2}} +\mathrm{1}\right\}\left({v}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\frac{\mathrm{1}}{\left\{\left({a}^{\mathrm{2}} +\mathrm{1}\right){v}^{\mathrm{2}} +\mathrm{1}\right\}\left({v}^{\mathrm{2}} +\mathrm{1}\right)}=\frac{{A}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right){v}^{\mathrm{2}} +\mathrm{1}}+\frac{{B}}{{v}^{\mathrm{2}} +\mathrm{1}} \\ $$$${A}\left({v}^{\mathrm{2}} +\mathrm{1}\right)+{B}\left({a}^{\mathrm{2}} +\mathrm{1}\right){v}^{\mathrm{2}} +{B}=\mathrm{1} \\ $$$$\left\{{A}+{B}\left({a}^{\mathrm{2}} +\mathrm{1}\right)\right\}{v}^{\mathrm{2}} +\left({A}+{B}\right)=\mathrm{1} \\ $$$$\Rightarrow{A}+{B}=\mathrm{1} \\ $$$$\Rightarrow{A}+{B}\left({a}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{B}=−\frac{\mathrm{1}}{{a}^{\mathrm{2}} } \\ $$$$\Rightarrow{A}=\frac{{a}^{\mathrm{2}} +\mathrm{1}}{{a}^{\mathrm{2}} } \\ $$$${I}={a}^{\mathrm{2}} \int\left[\frac{{a}^{\mathrm{2}} +\mathrm{1}}{{a}^{\mathrm{2}} \left\{\left({a}^{\mathrm{2}} +\mathrm{1}\right){v}^{\mathrm{2}} +\mathrm{1}\right\}}−\frac{\mathrm{1}}{{a}^{\mathrm{2}} \left({v}^{\mathrm{2}} +\mathrm{1}\right)}\right]{dv} \\ $$$${I}=\int\left[\frac{{a}^{\mathrm{2}} +\mathrm{1}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right){v}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{1}}{{v}^{\mathrm{2}} +\mathrm{1}}\right]{dv} \\ $$$${I}=\left({a}^{\mathrm{2}} +\mathrm{1}\right)\int\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right){v}^{\mathrm{2}} +\mathrm{1}}{dv}−\int\frac{\mathrm{1}}{{v}^{\mathrm{2}} +\mathrm{1}}{dv} \\ $$$${I}=\left({a}^{\mathrm{2}} +\mathrm{1}\right){I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\int\frac{\mathrm{1}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right){v}^{\mathrm{2}} +\mathrm{1}}{dv} \\ $$$${let}\:{p}=\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}{v} \\ $$$${dp}=\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\:{dv} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\int\frac{{dp}}{{p}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\mathrm{tan}^{−\mathrm{1}} {p} \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}{v}\right) \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{tan}\:\theta\right) \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\:\frac{\mathrm{sin}\:\theta}{\:\sqrt{\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \:\theta}}\right) \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\:\frac{\frac{{u}}{{a}}}{\:\sqrt{\mathrm{1}−\frac{{u}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}}\right) \\ $$$${I}_{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\:\frac{{u}}{\:\sqrt{{a}^{\mathrm{2}} −{u}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow{I}_{\mathrm{1}} =\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{tan}\:{x}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{tan}^{\mathrm{2}} \:{x}}}\right) \\ $$$${I}_{\mathrm{2}} =\int\frac{\mathrm{1}}{{v}^{\mathrm{2}} +\mathrm{1}}{dv}=\mathrm{tan}^{−\mathrm{1}} {v}=\theta=\mathrm{sin}^{−\mathrm{1}} \frac{{u}}{{a}} \\ $$$$\Rightarrow{I}_{\mathrm{2}} =\mathrm{sin}^{−\mathrm{1}} \frac{{u}}{{a}}=\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:{x}}{{a}}\right) \\ $$$${I}=\left({a}^{\mathrm{2}} +\mathrm{1}\right){I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$$\Rightarrow{I}=\left(\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right)\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\:\mathrm{tan}\:{x}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{tan}^{\mathrm{2}} \:{x}}}\right)−\mathrm{sin}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:{x}}{{a}}\right)+{C} \\ $$
Commented by malwaan last updated on 10/Oct/18
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by ajfour last updated on 10/Oct/18
$${Thanks}\:{a}\:{lot}\:{Sir}! \\ $$
Answered by ajfour last updated on 10/Oct/18
![let tan x = asin ψ ⇒ dx= ((acos ψdψ)/(1+a^2 sin^2 ψ)) I=∫((a^2 cos^2 ψ)/(1+a^2 sin^2 ψ))dψ let tan ψ = t ⇒ dψ=(dt/(1+t^2 )) I=∫((a^2 ((1/(1+t^2 )))(dt/((1+t^2 ))))/(1+(a^2 /((1+(1/t^2 )))))) = ∫((a^2 dt)/((1+t^2 )^2 +a^2 t^2 (1+t^2 ))) ............................................... = ∫((a^2 dt)/((a^2 +1)t^4 +(a^2 +2)t^2 +1)) = (a^2 /( (√(a^2 +1))))∫(((1/t^2 ))/( (√(a^2 +1))t^2 +(1/( (√(a^2 +1))t^2 ))+((a^2 +2)/( (√(a^2 +1))))))dt let t(a^2 +1)^(1/4) =bt=z ⇒ bdt = dz ⇒ I = ((b^4 −1)/b^2 )∫(((b^2 /z^2 )((dz/b)))/((z+(1/z))^2 +((b^4 +1)/b^2 )−2)) = ((b^4 −1)/(2b))∫(((1+(1/z^2 ))−(1−(1/z^2 )))/((z+(1/z))^2 +(((b^2 −1)/b))^2 )) dz = ((b^4 −1)/(2b))∫(((1+(1/z^2 ))dz)/((z−(1/z))^2 +(((b^2 +1)/b))^2 )) −((b^4 −1)/(2b))∫(((1−(1/z^2 ))dz)/((z+(1/z))^2 +(((b^2 −1)/b))^2 )) I = ((b^4 −1)/(2b))[(b/(b^2 +1))tan^(−1) (((z−(1/z))/((b^2 +1)/b)))−(b/(b^2 −1))tan^(−1) (((z+(1/z))/((b^2 −1)/b)))]+c And since bt=z , t=tan 𝛙 asin 𝛙 = tan x ⇒ tan 𝛙 = z/b ((az)/( (√(z^2 +b^2 )))) = tan x ∴ z^2 = ((b^2 tan^2 x)/(a^2 −tan^2 x)) ⇒ z = (((a^2 +1)^(1/4) tan x)/( (√(a^2 −tan^2 x)))) ; b=(a^2 +1)^(1/4) I = ((b^4 −1)/(2b))[(b/(b^2 +1))tan^(−1) (((z−(1/z))/((b^2 +1)/b)))−(b/(b^2 −1))tan^(−1) (((z+(1/z))/((b^2 −1)/b)))]+c . I= ((b^2 −1)/2)tan^(−1) ((b(z^2 −1))/(z(b^2 +1))) −((b^2 +1)/2)tan^(−1) ((b(z^2 +1))/(z(b^2 −1))) .](https://www.tinkutara.com/question/Q45219.png)
$${let}\:\:\mathrm{tan}\:{x}\:=\:{a}\mathrm{sin}\:\psi \\ $$$$\Rightarrow\:\:{dx}=\:\frac{{a}\mathrm{cos}\:\psi{d}\psi}{\mathrm{1}+{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \psi} \\ $$$${I}=\int\frac{{a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \psi}{\mathrm{1}+{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \psi}{d}\psi \\ $$$${let}\:\:\mathrm{tan}\:\psi\:=\:{t}\:\:\Rightarrow\:{d}\psi=\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${I}=\int\frac{{a}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)}}{\mathrm{1}+\frac{{a}^{\mathrm{2}} }{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}} \\ $$$$\:\:=\:\int\frac{{a}^{\mathrm{2}} {dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} +{a}^{\mathrm{2}} {t}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)} \\ $$$$……………………………………….. \\ $$$$\:\:=\:\int\frac{{a}^{\mathrm{2}} {dt}}{\left({a}^{\mathrm{2}} +\mathrm{1}\right){t}^{\mathrm{4}} +\left({a}^{\mathrm{2}} +\mathrm{2}\right){t}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\:\:=\:\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}\int\frac{\left(\mathrm{1}/{t}^{\mathrm{2}} \right)}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}{t}^{\mathrm{2}} +\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}{t}^{\mathrm{2}} }+\frac{{a}^{\mathrm{2}} +\mathrm{2}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}}{dt} \\ $$$$\:\:{let}\:\:{t}\left({a}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} ={bt}={z} \\ $$$$\Rightarrow\:\:\:\:{bdt}\:=\:{dz} \\ $$$$\Rightarrow\:{I}\:=\:\frac{{b}^{\mathrm{4}} −\mathrm{1}}{{b}^{\mathrm{2}} }\int\frac{\frac{{b}^{\mathrm{2}} }{{z}^{\mathrm{2}} }\left(\frac{{dz}}{{b}}\right)}{\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} +\frac{{b}^{\mathrm{4}} +\mathrm{1}}{{b}^{\mathrm{2}} }−\mathrm{2}} \\ $$$$\:\:\:=\:\frac{{b}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}{b}}\int\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right)}{\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} +\left(\frac{{b}^{\mathrm{2}} −\mathrm{1}}{{b}}\right)^{\mathrm{2}} }\:{dz} \\ $$$$\:\:=\:\frac{{b}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}{b}}\int\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right){dz}}{\left({z}−\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} +\left(\frac{{b}^{\mathrm{2}} +\mathrm{1}}{{b}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:−\frac{{b}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}{b}}\int\frac{\left(\mathrm{1}−\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right){dz}}{\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} +\left(\frac{{b}^{\mathrm{2}} −\mathrm{1}}{{b}}\right)^{\mathrm{2}} } \\ $$$$\:{I}\:=\:\frac{{b}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}{b}}\left[\frac{{b}}{{b}^{\mathrm{2}} +\mathrm{1}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{z}−\frac{\mathrm{1}}{{z}}}{\frac{{b}^{\mathrm{2}} +\mathrm{1}}{{b}}}\right)−\frac{{b}}{{b}^{\mathrm{2}} −\mathrm{1}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{z}+\frac{\mathrm{1}}{{z}}}{\frac{{b}^{\mathrm{2}} −\mathrm{1}}{{b}}}\right)\right]+{c} \\ $$$${And}\:{since}\:\:\boldsymbol{{bt}}=\boldsymbol{{z}}\:,\:\boldsymbol{{t}}=\mathrm{tan}\:\boldsymbol{\psi} \\ $$$$\:\:\:\:\:\boldsymbol{{a}}\mathrm{sin}\:\boldsymbol{\psi}\:=\:\mathrm{tan}\:\boldsymbol{{x}} \\ $$$$\Rightarrow\:\:\mathrm{tan}\:\boldsymbol{\psi}\:=\:\boldsymbol{{z}}/\boldsymbol{{b}} \\ $$$$\:\:\:\:\:\:\frac{\boldsymbol{{az}}}{\:\sqrt{\boldsymbol{{z}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} }}\:=\:\mathrm{tan}\:\boldsymbol{{x}} \\ $$$$\therefore\:\:\:\boldsymbol{{z}}^{\mathrm{2}} \:=\:\frac{\boldsymbol{{b}}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \boldsymbol{{x}}}{\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{tan}\:^{\mathrm{2}} \boldsymbol{{x}}} \\ $$$$\Rightarrow\:\:\boldsymbol{{z}}\:=\:\frac{\left(\boldsymbol{{a}}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} \mathrm{tan}\:\boldsymbol{{x}}}{\:\sqrt{\boldsymbol{{a}}^{\mathrm{2}} −\mathrm{tan}\:^{\mathrm{2}} \boldsymbol{{x}}}}\:\:\:;\:\boldsymbol{{b}}=\left(\boldsymbol{{a}}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{1}/\mathrm{4}} \\ $$$$\:{I}\:=\:\frac{{b}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}{b}}\left[\frac{{b}}{{b}^{\mathrm{2}} +\mathrm{1}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{z}−\frac{\mathrm{1}}{{z}}}{\frac{{b}^{\mathrm{2}} +\mathrm{1}}{{b}}}\right)−\frac{{b}}{{b}^{\mathrm{2}} −\mathrm{1}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{z}+\frac{\mathrm{1}}{{z}}}{\frac{{b}^{\mathrm{2}} −\mathrm{1}}{{b}}}\right)\right]+{c}\:\:. \\ $$$$\:{I}=\:\frac{{b}^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \frac{{b}\left({z}^{\mathrm{2}} −\mathrm{1}\right)}{{z}\left({b}^{\mathrm{2}} +\mathrm{1}\right)}\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:−\frac{{b}^{\mathrm{2}} +\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \frac{{b}\left({z}^{\mathrm{2}} +\mathrm{1}\right)}{{z}\left({b}^{\mathrm{2}} −\mathrm{1}\right)}\:\:\:\:. \\ $$
Commented by MrW3 last updated on 10/Oct/18
$${thanks}\:{sir}! \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18
$$\int\frac{{a}^{\mathrm{2}} −{tan}^{\mathrm{2}} {x}}{\:\sqrt{{a}^{\mathrm{2}} −{tan}^{\mathrm{2}} {x}}}{dx} \\ $$$${a}^{\mathrm{2}} \int\frac{{dx}}{\:\sqrt{{a}^{\mathrm{2}} −{tan}^{\mathrm{2}} {x}}}−\int\frac{{sec}^{\mathrm{2}} {x}−\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} −{tan}^{\mathrm{2}} {x}}}{dx} \\ $$$$\left({a}^{\mathrm{2}} +\mathrm{1}\right)\int\frac{{dx}}{\:\sqrt{{a}^{\mathrm{2}} −{tan}^{\mathrm{2}} {x}}}−\int\frac{{sec}^{\mathrm{2}} {x}\:{dx}}{\:\sqrt{{a}^{\mathrm{2}} −{tan}^{\mathrm{2}} {x}}} \\ $$$${I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\left({a}^{\mathrm{2}} +\mathrm{1}\right)\int\frac{{cosxdx}}{\:\sqrt{{a}^{\mathrm{2}} {cos}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {x}}} \\ $$$${t}={sinx}\:\:\:{dt}={cosx}\:{dx} \\ $$$$\left({a}^{\mathrm{2}} +\mathrm{1}\right)\int\frac{{dt}}{\:\sqrt{{a}^{\mathrm{2}} \left(\mathrm{1}−{t}^{\mathrm{2}} \right)−{t}^{\mathrm{2}} }} \\ $$$$\left({a}^{\mathrm{2}} +\mathrm{1}\right)\int\frac{{dt}}{\:\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} \left({a}^{\mathrm{2}} +\mathrm{1}\right)}} \\ $$$$\left({a}^{\mathrm{2}} +\mathrm{1}\right)\int\frac{{dt}}{\:\sqrt{\left({a}^{\mathrm{2}} +\mathrm{1}\right)\left(\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{1}}\right)−{t}^{\mathrm{2}} }} \\ $$$$\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\:\int\frac{{dt}}{\:\sqrt{\left(\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{1}}\right)^{} −{t}^{\mathrm{2}} }} \\ $$$$\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\:×{sin}^{−\mathrm{1}} \left(\frac{{t}}{\:\sqrt{\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{1}}}\:}\right) \\ $$$$\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\:×{sin}^{−\mathrm{1}} \left(\frac{{sinx}}{\:\sqrt{\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{1}}}\:}\right)+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int\frac{{sec}^{\mathrm{2}} {xdx}}{\:\sqrt{{a}^{\mathrm{2}} −{tan}^{\mathrm{2}} {x}}}{dx} \\ $$$${k}={tanx}\:\:\:{dk}={sec}^{\mathrm{2}} {xdx} \\ $$$$\int\frac{{dk}}{\:\sqrt{{a}^{\mathrm{2}} −{k}^{\mathrm{2}} }\:} \\ $$$${sin}^{−\mathrm{1}} \left(\frac{{k}}{{a}}\right)+{c}_{\mathrm{2}} \\ $$$${sin}^{−\mathrm{1}} \left(\frac{{tanx}}{{a}}\right)+{c}_{\mathrm{2}} \\ $$$$\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\:×{sin}^{−\mathrm{1}} \left(\frac{{sinx}}{\:\sqrt{\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{2}} +\mathrm{1}}}}\right)−{sin}^{−\mathrm{1}} \left(\frac{{tanx}}{{a}}\right)+{c} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18
$${pls}\:{check} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18
$${most}\:{welcome}\:{sir}… \\ $$
Commented by MrW3 last updated on 11/Oct/18
$${sir},\:{your}\:{answer}\:{is}\:{correct},\:{it}\:{can}\:{be} \\ $$$${transformed}\:{to}\:{the}\:{same}\:{form}\:{as}\:{mine}. \\ $$$${there}\:{is}\:{one}\:{strange}\:{thing}: \\ $$$${both}\:{in}\:{the}\:{result}\:{from}\:{me}\:{and}\:{from} \\ $$$${ajfour}\:{sir}\:{we}\:{have}\:{a}\:{limitation}\:{that} \\ $$$${is}\:\mathrm{tan}\:{x}\neq{a},\:{but}\:{in}\:{your}\:{result}\:{there} \\ $$$${is}\:{no}\:{limitation}.\:{but}\:{from}\:{the}\:{point}\:{of}\:{view} \\ $$$${of}\:{physic}\:{there}\:{should}\:{be}\:{this}\:{limitation}. \\ $$$${can}\:{you}\:{explain}\:{it}? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18
$${sir}\:{give}\:{me}\:{time}\:{to}\:{go}\:{through}\:{the}\:{problem}\:{of}\: \\ $$$${physics}\:{both}\:{of}\:{you}\:{sir}\:{solved}… \\ $$
Commented by MrW3 last updated on 11/Oct/18
$${thank}\:{you}\:{sir}. \\ $$$${it}'{s}\:{about}\:{the}\:{question}\:{if}\:{x}\:{can}\:{be}\:{all} \\ $$$${values}\:{from}\:\mathrm{0}\:{to}\:\frac{\pi}{\mathrm{2}}. \\ $$
Commented by MrW3 last updated on 11/Oct/18
$${thank}\:{you}\:{sir}! \\ $$