Question Number 187408 by MathsFan last updated on 17/Feb/23
$$\:\int\frac{\boldsymbol{\mathrm{sec}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}{\:\sqrt{\mathrm{1}−\boldsymbol{\mathrm{tan}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}}}\boldsymbol{\mathrm{dx}} \\ $$
Answered by horsebrand11 last updated on 17/Feb/23
$$\:=\:\int\:\frac{{d}\left(\mathrm{tan}\:{x}\right)}{\:\sqrt{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}}}\:;\:{u}=\mathrm{tan}\:{x} \\ $$$$=\int\:\frac{{du}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}\:;\:{u}=\mathrm{sin}\:{t} \\ $$$$=\int\frac{\mathrm{cos}\:{t}}{\:\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} {t}}}\:{dt}={t}\:+{C} \\ $$$$=\mathrm{arcsin}\:\left(\mathrm{tan}\:{x}\right)\:+\:{C} \\ $$
Commented by MJS_new last updated on 17/Feb/23
$$\mathrm{arcsin}\:\mathrm{tan}\:{x} \\ $$$$\mathrm{check}\:\mathrm{it}! \\ $$
Commented by horsebrand11 last updated on 17/Feb/23
$${yes}\:{thanks} \\ $$
Answered by MJS_new last updated on 17/Feb/23
![I love complicated paths ∫((sec^2 x)/( (√(1−tan^2 x))))dx=∫(dx/(cos x (√(1−2sin^2 x))))= [t=(((√2)sin x)/( (√(1−2sin^2 x)))) → dx=(((1−2sin^2 x)^(3/2) )/( (√2)cos x))] =(√2)∫(dt/(t^2 +2))=arctan (t/( (√2))) = =arctan ((sin x)/( (√(1−2sin^2 x)))) =arcsin tan x +C](https://www.tinkutara.com/question/Q187414.png)
$$\mathrm{I}\:\mathrm{love}\:\mathrm{complicated}\:\mathrm{paths} \\ $$$$\int\frac{\mathrm{sec}^{\mathrm{2}} \:{x}}{\:\sqrt{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{x}}}{dx}=\int\frac{{dx}}{\mathrm{cos}\:{x}\:\sqrt{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{x}}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{2}}\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{x}}}\:\rightarrow\:{dx}=\frac{\left(\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{x}\right)^{\mathrm{3}/\mathrm{2}} }{\:\sqrt{\mathrm{2}}\mathrm{cos}\:{x}}\right] \\ $$$$=\sqrt{\mathrm{2}}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}}=\mathrm{arctan}\:\frac{{t}}{\:\sqrt{\mathrm{2}}}\:= \\ $$$$=\mathrm{arctan}\:\frac{\mathrm{sin}\:{x}}{\:\sqrt{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{x}}}\:=\mathrm{arcsin}\:\mathrm{tan}\:{x}\:+{C} \\ $$
Commented by MathsFan last updated on 17/Feb/23
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$$$\mathrm{i}\:\mathrm{appreciate}\:\mathrm{your}\:\mathrm{effort} \\ $$