Question Number 69044 by mathmax by abdo last updated on 18/Sep/19
$$\left.\mathrm{1}\right)\:{let}\:{f}\left({a}\right)\:=\int\sqrt{{a}+{tanx}}{dx}\:\:{with}\:{a}>\mathrm{0} \\ $$$${find}\:{a}\:{explicit}\:{form}\:{for}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{find}\:{also}\:{g}\left({a}\right)\:=\int\:\:\frac{{dx}}{\:\sqrt{{a}+{tanx}}} \\ $$$$\left.\mathrm{3}\right){calculate}\:\int\sqrt{\mathrm{2}+{tanx}}{dx}\:{and}\:\int\:\:\frac{{dx}}{\:\sqrt{\mathrm{2}+{tanx}}} \\ $$
Answered by mind is power last updated on 18/Sep/19
![u=(√(a+tgx)) x=arctg(u^2 −a) dx=((2udu)/(1+(u^2 −a)^2 )) ∫(√(a+tan(x)))dx=∫((2u^2 du)/((u^2 −a)^2 +1)) =∫((2u^2 du)/(u^4 −2au^2 +a^2 +1))=∫((2u^2 du)/((u^2 +(√(a^2 +1)))^2 −(2a+2(√(a^2 +1)))u^2 )) =∫((2u^2 du)/((u^2 −(√((2a+2(√(a^2 +1)))))u+(√(a^2 +1)))(u^2 +(√(2a+(√(a^2 +1))))u+(√(a^2 +1))))) β=(√(2a+2(√(a^2 +1)))) η=(√(a^2 +1)) ∫((2u^2 )/((u^2 −βx+η)(x^2 +βx+η)))=((ax+b)/(x^2 −βx+η))+((cx+d)/(x^2 +βx+η)) a+c=0 b+d=0 β(a−c)+d+b=2 η(c+a)+β(−d+b)=0 ⇒a=−c b=−d ,−2cβ=2 b=d ,c=−(1/β),a=(1/β),b=d=0 ∫(x/(β(x^2 −βx+η)))dx−∫(dx/(β(x^2 +βx+η))) =(1/β)∫(dx/((x−(β/2))^2 +η−(β^2 /4)))−(1/β)∫(dx/((x+(β/2))^2 +η−(β^2 /4))) =(1/(β(√(η−(β^2 /4)))))tan^(−1) (((x−(β/2))/( (√(η−(β^2 /4))))))−(1/(β(√(η−(β^2 /4)))))tan^(−1) (((x+(β/2))/( (√(η−(β^2 /4))))))+c ∫(√(a+tan(x)))dx=(1/(β(√(η−(β^2 /4)))))[tan^(−1) ((((√(a+tg(x)))−(β/2))/( (√(η−(β^2 /4))))))−tan^(−1) ((((√(a+tg(x)))+(β/2))/( (√(η−(β^2 /4))))))+c f(a)=(1/( (√(2a+2(√(a^2 +1))(√(((√(a^2 +1))−a)/2))))))[tan^(−1) ((((√(a+tan (x)))−((√(2a+2(√(a^2 +1))))/2))/( (√(((√(a^2 +1))−a)/2)))))−tan^(−1) ((((√(a+tan(x)))+((√(2a+2(√(a^2 +1))))/2))/( (√(((√(a^2 +1))−a)/2)))))] f′(a)=∫(d/da)((√(a+tan (x)))dx)=(1/2)∫(dx/( (√(a+tan(x))))) ⇒∫(dx/( (√(a+tan(x)))))=2f′(a)′′verry long expression′′ 3)∫(√(2+tan (x)))dx=f(2) ∫(dx/( (√(2+tg(x)))))=2f′(2)](https://www.tinkutara.com/question/Q69050.png)
$${u}=\sqrt{{a}+{tgx}} \\ $$$${x}={arctg}\left({u}^{\mathrm{2}} −{a}\right) \\ $$$${dx}=\frac{\mathrm{2}{udu}}{\mathrm{1}+\left({u}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} } \\ $$$$\int\sqrt{{a}+{tan}\left({x}\right)}{dx}=\int\frac{\mathrm{2}{u}^{\mathrm{2}} {du}}{\left({u}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\int\frac{\mathrm{2}{u}^{\mathrm{2}} {du}}{{u}^{\mathrm{4}} −\mathrm{2}{au}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{1}}=\int\frac{\mathrm{2}{u}^{\mathrm{2}} {du}}{\left({u}^{\mathrm{2}} +\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} −\left(\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right){u}^{\mathrm{2}} } \\ $$$$=\int\frac{\mathrm{2}{u}^{\mathrm{2}} {du}}{\left({u}^{\mathrm{2}} −\sqrt{\left(\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right)}{u}+\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right)\left({u}^{\mathrm{2}} +\sqrt{\mathrm{2}{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}{u}+\sqrt{\left.{a}^{\mathrm{2}} +\mathrm{1}\right)}\right.} \\ $$$$\beta=\sqrt{\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\eta=\sqrt{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int\frac{\mathrm{2}{u}^{\mathrm{2}} }{\left({u}^{\mathrm{2}} −\beta{x}+\eta\right)\left({x}^{\mathrm{2}} +\beta{x}+\eta\right)}=\frac{{ax}+{b}}{{x}^{\mathrm{2}} −\beta{x}+\eta}+\frac{{cx}+{d}}{{x}^{\mathrm{2}} +\beta{x}+\eta} \\ $$$${a}+{c}=\mathrm{0} \\ $$$${b}+{d}=\mathrm{0} \\ $$$$\beta\left({a}−{c}\right)+{d}+{b}=\mathrm{2} \\ $$$$\eta\left({c}+{a}\right)+\beta\left(−{d}+{b}\right)=\mathrm{0} \\ $$$$\Rightarrow{a}=−{c}\:\:\:\:\:{b}=−{d}\:\:,−\mathrm{2}{c}\beta=\mathrm{2} \\ $$$${b}={d}\:\:\:,{c}=−\frac{\mathrm{1}}{\beta},{a}=\frac{\mathrm{1}}{\beta},{b}={d}=\mathrm{0} \\ $$$$\int\frac{{x}}{\beta\left({x}^{\mathrm{2}} −\beta{x}+\eta\right)}{dx}−\int\frac{{dx}}{\beta\left({x}^{\mathrm{2}} +\beta{x}+\eta\right)} \\ $$$$=\frac{\mathrm{1}}{\beta}\int\frac{{dx}}{\left({x}−\frac{\beta}{\mathrm{2}}\right)^{\mathrm{2}} +\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}−\frac{\mathrm{1}}{\beta}\int\frac{{dx}}{\left({x}+\frac{\beta}{\mathrm{2}}\right)^{\mathrm{2}} +\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}} \\ $$$$=\frac{\mathrm{1}}{\beta\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\beta}{\mathrm{2}}}{\:\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\right)−\frac{\mathrm{1}}{\beta\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}+\frac{\beta}{\mathrm{2}}}{\:\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\right)+{c} \\ $$$$\int\sqrt{{a}+{tan}\left({x}\right)}{dx}=\frac{\mathrm{1}}{\beta\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{a}+{tg}\left({x}\right)}−\frac{\beta}{\mathrm{2}}}{\:\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{a}+{tg}\left({x}\right)}+\frac{\beta}{\mathrm{2}}}{\:\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\right)+{c}\right. \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\sqrt{\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}−{a}}{\mathrm{2}}}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{a}+\mathrm{tan}\:\left({x}\right)}−\frac{\sqrt{\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}}{\mathrm{2}}}{\:\sqrt{\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}−{a}}{\mathrm{2}}}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{a}+{tan}\left({x}\right)}+\frac{\sqrt{\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}}{\mathrm{2}}}{\:\sqrt{\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}−{a}}{\mathrm{2}}}}\right)\right] \\ $$$${f}'\left({a}\right)=\int\frac{{d}}{{da}}\left(\sqrt{{a}+\mathrm{tan}\:\left({x}\right)}{dx}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\:\sqrt{{a}+{tan}\left({x}\right)}} \\ $$$$\Rightarrow\int\frac{{dx}}{\:\sqrt{{a}+{tan}\left({x}\right)}}=\mathrm{2}{f}'\left({a}\right)''{verry}\:{long}\:{expression}'' \\ $$$$\left.\mathrm{3}\right)\int\sqrt{\mathrm{2}+\mathrm{tan}\:\left({x}\right)}{dx}={f}\left(\mathrm{2}\right) \\ $$$$\int\frac{{dx}}{\:\sqrt{\mathrm{2}+{tg}\left({x}\right)}}=\mathrm{2}{f}'\left(\mathrm{2}\right) \\ $$$$ \\ $$
Commented by turbo msup by abdo last updated on 19/Sep/19
$${thank}\:{you}\:{sir}. \\ $$
Commented by mind is power last updated on 19/Sep/19
$${y}'{re}\:{welcom} \\ $$