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Question Number 59344 by rahul 19 last updated on 08/May/19
∫ e^x (((1−sin x)/(1−cos x)))dx = ?
$$\int\:{e}^{{x}} \left(\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}\right){dx}\:=\:? \\ $$
Answered by MJS last updated on 08/May/19
the trick is this: ((1−sin x)/(1−cos x))=(1/(1−cos x))−((sin x)/(1−cos x))=(1/(2sin^2 (x/2)))−(1/(tan (x/2)))= =(1/2)csc^2 (x/2) −cot (x/2) so we have ∫e^x (((1−sin x)/(1−cos x)))dx=(1/2)∫e^x csc^2 (x/2) dx−∫e^x cot (x/2) dx now solve the 2^(nd) one by parts ∫f′g=fg−∫fg′ f′=f=e^x g=cot (x/2) ⇒ g′=−((csc^2 x)/2) ∫e^x cot (x/2) dx=e^x cot (x/2) +(1/2)∫e^x csc^2 (x/2) dx ⇒ ∫e^x (((1−sin x)/(1−cos x)))dx=−e^x cot (x/2) +C
$$\mathrm{the}\:\mathrm{trick}\:\mathrm{is}\:\mathrm{this}: \\ $$$$\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{cos}\:{x}}−\frac{\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}=\frac{\mathrm{1}}{\mathrm{2sin}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{tan}\:\frac{{x}}{\mathrm{2}}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{csc}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\:−\mathrm{cot}\:\frac{{x}}{\mathrm{2}} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{have} \\ $$$$\int\mathrm{e}^{{x}} \left(\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{e}^{{x}} \mathrm{csc}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\:{dx}−\int\mathrm{e}^{{x}} \mathrm{cot}\:\frac{{x}}{\mathrm{2}}\:{dx} \\ $$$$\mathrm{now}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one}\:\mathrm{by}\:\mathrm{parts} \\ $$$$\int{f}'{g}={fg}−\int{fg}' \\ $$$${f}'={f}=\mathrm{e}^{{x}} \\ $$$${g}=\mathrm{cot}\:\frac{{x}}{\mathrm{2}}\:\Rightarrow\:{g}'=−\frac{\mathrm{csc}^{\mathrm{2}} \:{x}}{\mathrm{2}} \\ $$$$\int\mathrm{e}^{{x}} \mathrm{cot}\:\frac{{x}}{\mathrm{2}}\:{dx}=\mathrm{e}^{{x}} \mathrm{cot}\:\frac{{x}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}\int\mathrm{e}^{{x}} \mathrm{csc}^{\mathrm{2}} \:\frac{{x}}{\mathrm{2}}\:{dx} \\ $$$$\Rightarrow \\ $$$$\int\mathrm{e}^{{x}} \left(\frac{\mathrm{1}−\mathrm{sin}\:{x}}{\mathrm{1}−\mathrm{cos}\:{x}}\right){dx}=−\mathrm{e}^{{x}} \mathrm{cot}\:\frac{{x}}{\mathrm{2}}\:+{C} \\ $$
Commented by rahul 19 last updated on 08/May/19
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$
Commented by MJS last updated on 08/May/19
you're welcome; please check my answers before using them because I'm a typo-lover ��
Commented by rahul 19 last updated on 08/May/19
well, I already have answers to almost all the Q., I post here! All I want is "trick" .��
Commented by malwaan last updated on 09/May/19
fantastic sir ! what is typo−lover?
$${fantastic}\:{sir}\:! \\ $$$${what}\:{is}\:{typo}−{lover}? \\ $$
Answered by tanmay last updated on 08/May/19
∫e^x (((1−2sin(x/2)cos(x/2))/(2sin^2 (x/2))))dx ∫(e^x ×((cosec^2 (x/2))/2)−e^x ×cot(x/2))dx (−1)∫e^x cot(x/2)−e^x ×((cosec^2 (x/2))/2)dx (−1)∫[cot(x/2)×(de^x /dx)+e^x ×(d/dx)(cot(x/2))]dx (−1)∫(d/dx)(e^x cot(x/2))dx −e^x cot(x/2)+c
$$\int{e}^{{x}} \left(\frac{\mathrm{1}−\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}\right){dx} \\ $$$$\int\left({e}^{{x}} ×\frac{{cosec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{2}}−{e}^{{x}} ×{cot}\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$\left(−\mathrm{1}\right)\int{e}^{{x}} {cot}\frac{{x}}{\mathrm{2}}−{e}^{{x}} ×\frac{{cosec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{\mathrm{2}}{dx} \\ $$$$\left(−\mathrm{1}\right)\int\left[{cot}\frac{{x}}{\mathrm{2}}×\frac{{de}^{{x}} }{{dx}}+{e}^{{x}} ×\frac{{d}}{{dx}}\left({cot}\frac{{x}}{\mathrm{2}}\right)\right]{dx} \\ $$$$\left(−\mathrm{1}\right)\int\frac{{d}}{{dx}}\left({e}^{{x}} {cot}\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$−{e}^{{x}} {cot}\frac{{x}}{\mathrm{2}}+{c} \\ $$$$ \\ $$
Commented by rahul 19 last updated on 09/May/19
thank you sir!

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