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
$${If}\:{it}'{s}\:{too}\:{blurry},\:{please}\:{mention}. \\ $$$${I}'{ll}\:{write}\:{them}\:{in}\:{LaTeX}\:{form}. \\ $$
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](https://www.tinkutara.com/question/Q214858.png)
$$\int\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \:{x}}{dx}=\int\frac{\mathrm{1}+\mathrm{cos}\:{x}}{\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \:{x}}{dx}= \\ $$$$=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{{dx}}{\:\sqrt{\mathrm{2}}−\mathrm{cos}\:{x}}−\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{{dx}}{\:\sqrt{\mathrm{2}}+\mathrm{cos}\:{x}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right] \\ $$$$=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{dt}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){t}^{\mathrm{2}} −\mathrm{1}+\sqrt{\mathrm{2}}}+\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{dt}}{\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){t}^{\mathrm{2}} −\mathrm{1}−\sqrt{\mathrm{2}}}= \\ $$$$=\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\left(\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){t}\right)\:−\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\left(\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){t}\right)\:= \\ $$$$==\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\left(\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\:−\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \:\left(\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)\mathrm{tan}\:\frac{{x}}{\mathrm{2}}\right)\:+{C} \\ $$