Question Number 166449 by cortano1 last updated on 20/Feb/22
$$\:\:\:\mathrm{C}\:=\:\int\:\frac{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{1}+\mathrm{sec}\:^{\mathrm{2}} \mathrm{x}}\:\mathrm{dx}\:=? \\ $$
Commented by cortano1 last updated on 20/Feb/22
$$\mathrm{oo}\:\mathrm{yes} \\ $$
Commented by cortano1 last updated on 20/Feb/22
Commented by MJS_new last updated on 20/Feb/22
$${u}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\:\Leftrightarrow\:{x}=\mathrm{2arctan}\:{u}\:\Rightarrow\:\mathrm{cos}^{\mathrm{2}} \:{x}\:=\left(\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$
Answered by MJS_new last updated on 20/Feb/22
![∫((1−tan^2 x)/(1+sec^2 x))dx=2∫dx−3∫(1/(1+cos^2 x))dx= [t=tan x → dx=cos^2 x dt] =2x−3∫(dt/(t^2 +2))= =2x−(3/( (√2)))arctan (t/( (√2))) = =2x−(3/( (√2)))arctan ((tan x)/( (√2))) +C](Q166459.png)
$$\int\frac{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} \:{x}}{\mathrm{1}+\mathrm{sec}^{\mathrm{2}} \:{x}}{dx}=\mathrm{2}\int{dx}−\mathrm{3}\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:{x}\:\rightarrow\:{dx}=\mathrm{cos}^{\mathrm{2}} \:{x}\:{dt}\right] \\ $$$$=\mathrm{2}{x}−\mathrm{3}\int\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{2}}= \\ $$$$=\mathrm{2}{x}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\mathrm{arctan}\:\frac{{t}}{\:\sqrt{\mathrm{2}}}\:= \\ $$$$=\mathrm{2}{x}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{2}}}\mathrm{arctan}\:\frac{\mathrm{tan}\:{x}}{\:\sqrt{\mathrm{2}}}\:+{C} \\ $$
Commented by peter frank last updated on 24/Feb/22
$$\mathrm{great} \\ $$