1-let-f-a-a-tanx-dx-with-a-gt-0-find-a-explicit-form-for-f-x-2-find-also-g-a-dx-a-tanx-3-calculate-2-tanx-dx-and-dx-2-tanx-

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
$${u}=\sqrt{{a}+{tgx}} \\$$$${x}={arctg}\left({u}^{\mathrm{2}} −{a}\right) \\$$$${dx}=\frac{\mathrm{2}{udu}}{\mathrm{1}+\left({u}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} } \\$$$$\int\sqrt{{a}+{tan}\left({x}\right)}{dx}=\int\frac{\mathrm{2}{u}^{\mathrm{2}} {du}}{\left({u}^{\mathrm{2}} −{a}\right)^{\mathrm{2}} +\mathrm{1}} \\$$$$=\int\frac{\mathrm{2}{u}^{\mathrm{2}} {du}}{{u}^{\mathrm{4}} −\mathrm{2}{au}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{1}}=\int\frac{\mathrm{2}{u}^{\mathrm{2}} {du}}{\left({u}^{\mathrm{2}} +\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right)^{\mathrm{2}} −\left(\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right){u}^{\mathrm{2}} } \\$$$$=\int\frac{\mathrm{2}{u}^{\mathrm{2}} {du}}{\left({u}^{\mathrm{2}} −\sqrt{\left(\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right)}{u}+\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\right)\left({u}^{\mathrm{2}} +\sqrt{\mathrm{2}{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}{u}+\sqrt{\left.{a}^{\mathrm{2}} +\mathrm{1}\right)}\right.} \\$$$$\beta=\sqrt{\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}} \\$$$$\eta=\sqrt{{a}^{\mathrm{2}} +\mathrm{1}} \\$$$$\int\frac{\mathrm{2}{u}^{\mathrm{2}} }{\left({u}^{\mathrm{2}} −\beta{x}+\eta\right)\left({x}^{\mathrm{2}} +\beta{x}+\eta\right)}=\frac{{ax}+{b}}{{x}^{\mathrm{2}} −\beta{x}+\eta}+\frac{{cx}+{d}}{{x}^{\mathrm{2}} +\beta{x}+\eta} \\$$$${a}+{c}=\mathrm{0} \\$$$${b}+{d}=\mathrm{0} \\$$$$\beta\left({a}−{c}\right)+{d}+{b}=\mathrm{2} \\$$$$\eta\left({c}+{a}\right)+\beta\left(−{d}+{b}\right)=\mathrm{0} \\$$$$\Rightarrow{a}=−{c}\:\:\:\:\:{b}=−{d}\:\:,−\mathrm{2}{c}\beta=\mathrm{2} \\$$$${b}={d}\:\:\:,{c}=−\frac{\mathrm{1}}{\beta},{a}=\frac{\mathrm{1}}{\beta},{b}={d}=\mathrm{0} \\$$$$\int\frac{{x}}{\beta\left({x}^{\mathrm{2}} −\beta{x}+\eta\right)}{dx}−\int\frac{{dx}}{\beta\left({x}^{\mathrm{2}} +\beta{x}+\eta\right)} \\$$$$=\frac{\mathrm{1}}{\beta}\int\frac{{dx}}{\left({x}−\frac{\beta}{\mathrm{2}}\right)^{\mathrm{2}} +\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}−\frac{\mathrm{1}}{\beta}\int\frac{{dx}}{\left({x}+\frac{\beta}{\mathrm{2}}\right)^{\mathrm{2}} +\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}} \\$$$$=\frac{\mathrm{1}}{\beta\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\beta}{\mathrm{2}}}{\:\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\right)−\frac{\mathrm{1}}{\beta\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{x}+\frac{\beta}{\mathrm{2}}}{\:\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\right)+{c} \\$$$$\int\sqrt{{a}+{tan}\left({x}\right)}{dx}=\frac{\mathrm{1}}{\beta\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{a}+{tg}\left({x}\right)}−\frac{\beta}{\mathrm{2}}}{\:\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{a}+{tg}\left({x}\right)}+\frac{\beta}{\mathrm{2}}}{\:\sqrt{\eta−\frac{\beta^{\mathrm{2}} }{\mathrm{4}}}}\right)+{c}\right. \\$$$${f}\left({a}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}\sqrt{\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}−{a}}{\mathrm{2}}}}}\left[\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{a}+\mathrm{tan}\:\left({x}\right)}−\frac{\sqrt{\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}}{\mathrm{2}}}{\:\sqrt{\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}−{a}}{\mathrm{2}}}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{a}+{tan}\left({x}\right)}+\frac{\sqrt{\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}}}{\mathrm{2}}}{\:\sqrt{\frac{\sqrt{{a}^{\mathrm{2}} +\mathrm{1}}−{a}}{\mathrm{2}}}}\right)\right] \\$$$${f}'\left({a}\right)=\int\frac{{d}}{{da}}\left(\sqrt{{a}+\mathrm{tan}\:\left({x}\right)}{dx}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dx}}{\:\sqrt{{a}+{tan}\left({x}\right)}} \\$$$$\Rightarrow\int\frac{{dx}}{\:\sqrt{{a}+{tan}\left({x}\right)}}=\mathrm{2}{f}'\left({a}\right)''{verry}\:{long}\:{expression}'' \\$$$$\left.\mathrm{3}\right)\int\sqrt{\mathrm{2}+\mathrm{tan}\:\left({x}\right)}{dx}={f}\left(\mathrm{2}\right) \\$$$$\int\frac{{dx}}{\:\sqrt{\mathrm{2}+{tg}\left({x}\right)}}=\mathrm{2}{f}'\left(\mathrm{2}\right) \\$$$$\\$$
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
$${y}'{re}\:{welcom} \\$$