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Question Number 212141 by MrGaster last updated on 03/Oct/24
∫((cos^2 x)/(sin x+cos x))dx.
$$ \\ $$$$\:\:\:\:\:\:\:\:\int\frac{\mathrm{cos}^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}{dx}. \\ $$$$ \\ $$
Answered by BHOOPENDRA last updated on 03/Oct/24
∫((cos^2 x)/(sin x+cos x)) dx =^(t=tan((x/2))) ∫ (((t−1)^2 (t+1)^2 )/((t^2 +1)^2 (t^2 −2t−1)))dt (1/2)∫(1/((t^2 −2t−1)))dt+(1/2)∫(1/((t^2 +1)))dt+∫(((t−1))/((t^2 +1)^2 )) dt ((ln (t−(√2)+1))/2^(3/2) ) −((ln (t+(√2)−1))/2^(3/2) )+(1/2)tan^(−1) (t)−(t/(2(t^2 +1)))−(1/(2(t^2 +1))) (x/2)+((ln (tan((x/2))−(√2)+1))/2^(3/2) )+((ln (tan ((x/2))+(√2)−1))/2^(3/2) ) −((tan ((x/2)))/(2(tan^2 ((x/2))+1))) −(1/(2(tan^2 ((x/2))+1)))+C
$$\int\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}\:{dx}\:\overset{{t}={tan}\left(\frac{{x}}{\mathrm{2}}\right)} {=}\: \\ $$$$\int\:\frac{\left({t}−\mathrm{1}\right)^{\mathrm{2}} \left({t}+\mathrm{1}\right)^{\mathrm{2}} }{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)}{dt} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} −\mathrm{2}{t}−\mathrm{1}\right)}{dt}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)}{dt}+\int\frac{\left({t}−\mathrm{1}\right)}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\:{dt} \\ $$$$\frac{\mathrm{ln}\:\left({t}−\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:−\frac{\mathrm{ln}\:\left({t}+\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} }+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \left({t}\right)−\frac{{t}}{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)}−\frac{\mathrm{1}}{\mathrm{2}\left({t}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\frac{{x}}{\mathrm{2}}+\frac{\mathrm{ln}\:\left(\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)−\sqrt{\mathrm{2}}+\mathrm{1}\right)}{\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} }+\frac{\mathrm{ln}\:\left(\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)+\sqrt{\mathrm{2}}−\mathrm{1}\right)}{\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} } \\ $$$$\:\:\:\:\:−\frac{\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}\left(\mathrm{tan}\:^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}\right)}\:−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{tan}^{\mathrm{2}} \:\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{1}\right)}+{C} \\ $$
Answered by Frix last updated on 03/Oct/24
∫((cos^2 x)/(cos x +sin x))dx= =∫(((cos x −sin x)/2)+(1/(2(cos x sin x)))dx= =((√2)/2)∫cos (x+(π/4)) dx+((√2)/4)∫(dx/(sin (x+(π/4))))= =((√2)/2)sin (x+(π/4)) +((√2)/4)ln ∣tan ((x/2)+(π/8))∣ +C
$$\int\frac{\mathrm{cos}^{\mathrm{2}} \:{x}}{\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}}{dx}= \\ $$$$=\int\left(\frac{\mathrm{cos}\:{x}\:−\mathrm{sin}\:{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{cos}\:{x}\:\mathrm{sin}\:{x}\right.}\right){dx}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\mathrm{cos}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\:{dx}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\int\frac{{dx}}{\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)}= \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\mathrm{sin}\:\left({x}+\frac{\pi}{\mathrm{4}}\right)\:+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\mathrm{ln}\:\mid\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}\right)\mid\:+{C} \\ $$

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