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Question Number 61056 by Tawa1 last updated on 28/May/19
∫ ((x + sin(x))/(1 + cos(x))) dx
$$\int\:\frac{\mathrm{x}\:+\:\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{1}\:+\:\mathrm{cos}\left(\mathrm{x}\right)}\:\mathrm{dx} \\ $$
Answered by perlman last updated on 28/May/19
cos(x)=2cos^2 ((x/2))−1 ((x+sin(x))/(1+cos(x)))=((x+sin(x))/(2cos^2 ((x/2))))=(x/(2cos^2 ((x/2))))+((2sin((x/2))cos((x/2)))/(2cos^2 ((x/2)))) =(x/2)(1+tan^2 ((x/2)))+((sin((x/2)))/(cos((x/2)))) ==>∫((x+sin(x))/(1+cos(x)))dx=∫(x/2)(1+tg^2 ((x/2)))dx+∫((sin((x/2)))/(cos((x/2))))dx ∫((sin((x/2)))/(cos((x/2))))dx=−2ln∣cos((x/2))∣ ∫(x/2)(1+tan^2 ((x/2)))dx=xtan((x/2))−∫tan((x/2))dx=xtan((x/2))+2ln∣cos((x/2))∣+c ∫((x+sin(x))/(cos(x)+1))dx=xtan((x/2))+c c constant....
$${cos}\left({x}\right)=\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)−\mathrm{1} \\ $$$$\frac{{x}+{sin}\left({x}\right)}{\mathrm{1}+{cos}\left({x}\right)}=\frac{{x}+{sin}\left({x}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}=\frac{{x}}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}+\frac{\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$=\frac{{x}}{\mathrm{2}}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)+\frac{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{{cos}\left(\frac{{x}}{\mathrm{2}}\right)} \\ $$$$==>\int\frac{{x}+{sin}\left({x}\right)}{\mathrm{1}+{cos}\left({x}\right)}{dx}=\int\frac{{x}}{\mathrm{2}}\left(\mathrm{1}+{tg}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right){dx}+\int\frac{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{{cos}\left(\frac{{x}}{\mathrm{2}}\right)}{dx} \\ $$$$\int\frac{{sin}\left(\frac{{x}}{\mathrm{2}}\right)}{{cos}\left(\frac{{x}}{\mathrm{2}}\right)}{dx}=−\mathrm{2}{ln}\mid{cos}\left(\frac{{x}}{\mathrm{2}}\right)\mid \\ $$$$\int\frac{{x}}{\mathrm{2}}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right){dx}={xtan}\left(\frac{{x}}{\mathrm{2}}\right)−\int{tan}\left(\frac{{x}}{\mathrm{2}}\right){dx}={xtan}\left(\frac{{x}}{\mathrm{2}}\right)+\mathrm{2}{ln}\mid{cos}\left(\frac{{x}}{\mathrm{2}}\right)\mid+{c} \\ $$$$\int\frac{{x}+{sin}\left({x}\right)}{{cos}\left({x}\right)+\mathrm{1}}{dx}={xtan}\left(\frac{{x}}{\mathrm{2}}\right)+{c}\:\:\:\:\:\:\:{c}\:{constant}…. \\ $$$$ \\ $$
Commented by Tawa1 last updated on 28/May/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by perlman last updated on 28/May/19
2 nd x+sin(x)=x(cos^2 ((x/2))+sin^2 ((x/2)))+2sin((x/2))cos((x/2)) 1+cos(x)=2cos^2 ((x/2)) ∫ ((x + sin(x))/(1 + cos(x))) dx=∫(x/2)(1+tan^2 ((x/2)))+tg((x/2))dx=∫d(xtan((x/2)))=xtan((x/2))+c
$$\mathrm{2}\:{nd} \\ $$$${x}+{sin}\left({x}\right)={x}\left({cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)+{sin}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)+\mathrm{2}{sin}\left(\frac{{x}}{\mathrm{2}}\right){cos}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$\mathrm{1}+{cos}\left({x}\right)=\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$\int\:\frac{\mathrm{x}\:+\:\mathrm{sin}\left(\mathrm{x}\right)}{\mathrm{1}\:+\:\mathrm{cos}\left(\mathrm{x}\right)}\:\mathrm{dx}=\int\frac{{x}}{\mathrm{2}}\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right)+{tg}\left(\frac{{x}}{\mathrm{2}}\right){dx}=\int{d}\left({xtan}\left(\frac{{x}}{\mathrm{2}}\right)\right)={xtan}\left(\frac{{x}}{\mathrm{2}}\right)+{c} \\ $$$$ \\ $$
Commented by Tawa1 last updated on 28/May/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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