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Question Number 6865 by Tawakalitu. last updated on 31/Jul/16

∫((sin(x) + cos(x))/(e^(−x) + sin(x))) dx

$$\int\frac{{sin}\left({x}\right)\:+\:{cos}\left({x}\right)}{{e}^{−{x}} \:+\:{sin}\left({x}\right)}\:{dx} \\ $$

Commented by Yozzii last updated on 31/Jul/16

sinhix=isinx sinx=((sinhix)/i)=((e^(ix) −e^(−ix) )/(2i)) cosx=coshix=((e^(ix) +e^(−ix) )/2) sinx+cosx=((e^(ix) −e^(−ix) +ie^(ix) +ie^(−ix) )/(2i))=((e^(ix) (1+i)+e^(−ix) (i−1))/(2i)) e^(−x) +sinx=((2ie^(−x) +e^(ix) −e^(−ix) )/(2i)) I=∫((sinx+cosx)/(e^(−x) +sinx))dx=∫((e^(ix) (i+1)+e^(−ix) (i−1))/(e^(ix) −e^(−ix) +2ie^(−x) ))dx I=∫(e^x /e^x )×((e^(ix) (i+1)+e^(−ix) (i−1))/(e^(ix) −e^(−ix) +2ie^(−x) ))dx=∫((e^(x(i+1)) (i+1)−e^(x(1−i)) (1−i))/(e^(x(i+1)) −e^(x(1−i)) +2i))dx Let u=e^(x(1+i)) −e^(x(1−i)) +2i. ⇒du={e^(x(1+i)) (i+1)−e^(x(1−i)) (1−i)}dx ⇒I=∫(du/u)=ln∣u∣+C I=ln∣e^(x(1+i)) −e^(x(1−i)) +2i∣+C u=e^x (e^(ix) −e^(−ix) )+2i u=e^x (2isinx)+2i u=2i(e^x sinx+1) ∴ I=ln∣2i(e^x sinx+1)∣+C I=ln2+ln∣e^x sinx+1∣+C ∫((sinx+cosx)/(e^(−x) +sinx))dx=ln∣e^x sinx+1∣+D D=constant Checking: (d/dx)ln(e^x sinx+1)=((e^x (cosx+sinx))/(e^x sinx+1))=((sinx+cosx)/(sinx+e^(−x) )) −−−−−−−−−−−−−−−−−−−−−−−− This answer suggests the folliwing approach: I=∫((sinx+cosx)/(e^(−x) +sinx))dx=∫((e^x (sinx+cosx))/(1+e^x sinx))dx. Let u=1+e^x sinx⇒du=e^x (cosx+sinx)dx I=∫(du/u)=ln∣u∣=ln∣e^x sinx+1∣+C.

$${sinhix}={isinx} \\ $$$${sinx}=\frac{{sinhix}}{{i}}=\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}} \\ $$$${cosx}={coshix}=\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}} \\ $$$${sinx}+{cosx}=\frac{{e}^{{ix}} −{e}^{−{ix}} +{ie}^{{ix}} +{ie}^{−{ix}} }{\mathrm{2}{i}}=\frac{{e}^{{ix}} \left(\mathrm{1}+{i}\right)+{e}^{−{ix}} \left({i}−\mathrm{1}\right)}{\mathrm{2}{i}} \\ $$$${e}^{−{x}} +{sinx}=\frac{\mathrm{2}{ie}^{−{x}} +{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}} \\ $$$${I}=\int\frac{{sinx}+{cosx}}{{e}^{−{x}} +{sinx}}{dx}=\int\frac{{e}^{{ix}} \left({i}+\mathrm{1}\right)+{e}^{−{ix}} \left({i}−\mathrm{1}\right)}{{e}^{{ix}} −{e}^{−{ix}} +\mathrm{2}{ie}^{−{x}} }{dx} \\ $$$${I}=\int\frac{{e}^{{x}} }{{e}^{{x}} }×\frac{{e}^{{ix}} \left({i}+\mathrm{1}\right)+{e}^{−{ix}} \left({i}−\mathrm{1}\right)}{{e}^{{ix}} −{e}^{−{ix}} +\mathrm{2}{ie}^{−{x}} }{dx}=\int\frac{{e}^{{x}\left({i}+\mathrm{1}\right)} \left({i}+\mathrm{1}\right)−{e}^{{x}\left(\mathrm{1}−{i}\right)} \left(\mathrm{1}−{i}\right)}{{e}^{{x}\left({i}+\mathrm{1}\right)} −{e}^{{x}\left(\mathrm{1}−{i}\right)} +\mathrm{2}{i}}{dx} \\ $$$${Let}\:{u}={e}^{{x}\left(\mathrm{1}+{i}\right)} −{e}^{{x}\left(\mathrm{1}−{i}\right)} +\mathrm{2}{i}. \\ $$$$\Rightarrow{du}=\left\{{e}^{{x}\left(\mathrm{1}+{i}\right)} \left({i}+\mathrm{1}\right)−{e}^{{x}\left(\mathrm{1}−{i}\right)} \left(\mathrm{1}−{i}\right)\right\}{dx} \\ $$$$\Rightarrow{I}=\int\frac{{du}}{{u}}={ln}\mid{u}\mid+{C} \\ $$$${I}={ln}\mid{e}^{{x}\left(\mathrm{1}+{i}\right)} −{e}^{{x}\left(\mathrm{1}−{i}\right)} +\mathrm{2}{i}\mid+{C} \\ $$$${u}={e}^{{x}} \left({e}^{{ix}} −{e}^{−{ix}} \right)+\mathrm{2}{i} \\ $$$${u}={e}^{{x}} \left(\mathrm{2}{isinx}\right)+\mathrm{2}{i} \\ $$$${u}=\mathrm{2}{i}\left({e}^{{x}} {sinx}+\mathrm{1}\right) \\ $$$$\therefore\:{I}={ln}\mid\mathrm{2}{i}\left({e}^{{x}} {sinx}+\mathrm{1}\right)\mid+{C} \\ $$$${I}={ln}\mathrm{2}+{ln}\mid{e}^{{x}} {sinx}+\mathrm{1}\mid+{C} \\ $$$$\int\frac{{sinx}+{cosx}}{{e}^{−{x}} +{sinx}}{dx}={ln}\mid{e}^{{x}} {sinx}+\mathrm{1}\mid+{D} \\ $$$${D}={constant} \\ $$$${Checking}: \\ $$$$\frac{{d}}{{dx}}{ln}\left({e}^{{x}} {sinx}+\mathrm{1}\right)=\frac{{e}^{{x}} \left({cosx}+{sinx}\right)}{{e}^{{x}} {sinx}+\mathrm{1}}=\frac{{sinx}+{cosx}}{{sinx}+{e}^{−{x}} } \\ $$$$−−−−−−−−−−−−−−−−−−−−−−−− \\ $$$${This}\:{answer}\:{suggests}\:{the}\:{folliwing}\:{approach}: \\ $$$${I}=\int\frac{{sinx}+{cosx}}{{e}^{−{x}} +{sinx}}{dx}=\int\frac{{e}^{{x}} \left({sinx}+{cosx}\right)}{\mathrm{1}+{e}^{{x}} {sinx}}{dx}. \\ $$$${Let}\:{u}=\mathrm{1}+{e}^{{x}} {sinx}\Rightarrow{du}={e}^{{x}} \left({cosx}+{sinx}\right){dx} \\ $$$${I}=\int\frac{{du}}{{u}}={ln}\mid{u}\mid={ln}\mid{e}^{{x}} {sinx}+\mathrm{1}\mid+{C}. \\ $$

Commented by Tawakalitu. last updated on 31/Jul/16

Wow ... I really appreciate your effort

$${Wow}\:...\:{I}\:{really}\:{appreciate}\:{your}\:{effort} \\ $$

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