Question Number 214853 by liuxinnan last updated on 21/Dec/24
$$\int\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} {x}}{dx}=? \\ $$
Commented by liuxinnan last updated on 21/Dec/24
$${I}\:{had}\:{long}\:{time}\:{not}\:{look}\:{through} \\ $$$${this}\:{app}\:,{I}\:{just}\:{find} \\ $$
Answered by TonyCWX08 last updated on 21/Dec/24
Commented by TonyCWX08 last updated on 21/Dec/24
$${If}\:{it}'{s}\:{too}\:{blurry},\:{please}\:{mention}. \\ $$$${I}'{ll}\:{write}\:{them}\:{in}\:{LaTeX}\:{form}. \\ $$
Commented by liuxinnan last updated on 25/Dec/24
$${thanks} \\ $$
Answered by Frix last updated on 21/Dec/24
![∫((1+cos x)/(1+sin^2 x))dx=∫((1+cos x)/(2−cos^2 x))dx= =((2+(√2))/4)∫(dx/( (√2)−cos x))−((2−(√2))/4)∫(dx/( (√2)+cos x))= [t=tan (x/2)] =((2+(√2))/2)∫(dt/((1+(√2))t^2 −1+(√2)))+((2−(√2))/2)∫(dt/((1−(√2))t^2 −1−(√2)))= =((2+(√2))/2)tan^(−1) ((1+(√2))t) −((2−(√2))/2)tan^(−1) ((1−(√2))t) = ==((2+(√2))/2)tan^(−1) ((1+(√2))tan (x/2)) −((2−(√2))/2)tan^(−1) ((1−(√2))tan (x/2)) +C](Q214858.png)
$$\int\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}=\int\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{{dx}}{\:\sqrt{\mathrm{2}}−\mathrm{cos}\:{x}}−\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{{dx}}{\:\sqrt{\mathrm{2}}+\mathrm{cos}\:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right] \\ $$$$=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{dt}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){t}^{\mathrm{2}} −\mathrm{1}+\sqrt{\mathrm{2}}}+\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{dt}}{\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){t}^{\mathrm{2}} −\mathrm{1}−\sqrt{\mathrm{2}}}= \\ $$$$=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\left(\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){t}\right)\:−\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\left(\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){t}\right)\:= \\ $$$$==\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\left(\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\:−\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\left(\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\:+{C} \\ $$
Commented by liuxinnan last updated on 25/Dec/24
$${thanks} \\ $$
Answered by MrGaster last updated on 22/Dec/24
$$\int\frac{\mathrm{1}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}}{dx}+\int\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}}{dx} \\ $$$$\int\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} {x}}{dx}\Rrightarrow\mathrm{sin}^{\mathrm{2}} {x}=\mathrm{1}−\mathrm{cos}\:{x} \\ $$$$\therefore\Rrightarrow\int\frac{\mathrm{1}}{\mathrm{2}−\mathrm{cos}^{\mathrm{2}} {x}}{dx}+\int\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}}{dx}\Rrightarrow\mathrm{2}−\mathrm{cos}^{\mathrm{2}} {x}=\mathrm{2}−\frac{\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} {x}}=\frac{\mathrm{2sec}^{\mathrm{2}} {x}−\mathrm{1}}{\mathrm{sec}^{\mathrm{2}} {x}} \\ $$$$\therefore\Rrightarrow\int\frac{\mathrm{sec}^{\mathrm{2}} {x}}{\mathrm{2sec}^{\mathrm{2}} {x}−\mathrm{1}}{dx}+\int\frac{\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}}{dx}\Rrightarrow\int\frac{\mathrm{sec}^{\mathrm{2}} {x}}{\mathrm{1}+\left(\sqrt{\mathrm{2}}\mathrm{tan}\:{x}\right)^{\mathrm{2}} }{dx}+\int\frac{{d}\left(\mathrm{sin}\:{x}\right)}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{arctan}\left(\sqrt{\mathrm{2}}\mathrm{tan}\:{x}\right)+\mathrm{arctan}\left(\mathrm{sin}\:{x}\right)+{C} \\ $$
Commented by liuxinnan last updated on 25/Dec/24
$${thanks} \\ $$