Question Number 44992 by manish09@gmail.com last updated on 07/Oct/18
$$\int\mathrm{e}^{\mathrm{x}} \left(\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}\right)\mathrm{dx}\:? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Oct/18
![∫e^x (((1−2sin(x/2)cos(x/2))/(2cos^2 (x/2)))) ∫e^x ((1/2)sec^2 (x/2)−tan(x/2))dx (1/2)∫e^x sec^2 (x/2)dx−∫e^x tan(x/2)dx (1/2)∫e^x sec^2 (x/2)dx−[tan(x/2)∫e^x −∫(1/2)sec^2 (x/2) e^x dx] ∫e^x sec^2 (x/2)dx− e^x tan(x/2)+c i think problem should be ∫e^x (((1+sinx)/(1+cosx)))dx pls check](https://www.tinkutara.com/question/Q45003.png)
$$\int{e}^{{x}} \left(\frac{\mathrm{1}−\mathrm{2}{sin}\frac{{x}}{\mathrm{2}}{cos}\frac{{x}}{\mathrm{2}}}{\mathrm{2}{cos}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}\right) \\ $$$$\int{e}^{{x}} \left(\frac{\mathrm{1}}{\mathrm{2}}{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−{tan}\frac{{x}}{\mathrm{2}}\right){dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{{x}} {sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}−\int{e}^{{x}} {tan}\frac{{x}}{\mathrm{2}}{dx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{{x}} {sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}−\left[{tan}\frac{{x}}{\mathrm{2}}\int{e}^{{x}} −\int\frac{\mathrm{1}}{\mathrm{2}}{sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\:{e}^{{x}} {dx}\right] \\ $$$$\int{e}^{{x}} {sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}{dx}−\:{e}^{{x}} {tan}\frac{{x}}{\mathrm{2}}+{c} \\ $$$${i}\:{think}\:{problem}\:{should}\:{be} \\ $$$$\int{e}^{{x}} \left(\frac{\mathrm{1}+{sinx}}{\mathrm{1}+{cosx}}\right){dx}\:{pls}\:{check} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Oct/18
$${wait}\:{pls}… \\ $$
Commented by manish09@gmail.com last updated on 07/Oct/18
$$\mathrm{no}\:\mathrm{sir}\:\:\mathrm{this}\:\mathrm{is}\:\mathrm{right} \\ $$
Commented by manish09@gmail.com last updated on 07/Oct/18
$$\mathrm{pls}\:\mathrm{sir}\:\mathrm{solv}\:\mathrm{this}\:\:\mathrm{que}. \\ $$
Answered by ajfour last updated on 07/Oct/18
![∫e^x [((f(x))/(1+cos x))+((f ′(x)(1+cos x)+f(x)sin x)/((1+cos x)^2 ))]dx =∫e^x [((1−sin x+cos x−sin xcos x)/((1+cos x)^2 ))]dx ⇒ f(x)+f(x)cos x+f ′(x)(1+cos x)+f(x)sin x = 1−sin x+cos x−sin xcos x ⇒f(x)(1+cos x+sin x)+f ′(x)(1+cos x) = 1−sin x+cos x−sin xcos x ⇒ f ′(x)+((1+cos x+sin x)/(1+cos x))f(x)=1−sin x e^x f(x)=∫e^(∫(1+cot (x/2))dx) (1−sin x)dx ⇒ e^x f(x)=(1/2)∫e^x (1−cos x)(1−sin x)dx .....](https://www.tinkutara.com/question/Q45013.png)
$$\int{e}^{{x}} \left[\frac{{f}\left({x}\right)}{\mathrm{1}+\mathrm{cos}\:{x}}+\frac{{f}\:'\left({x}\right)\left(\mathrm{1}+\mathrm{cos}\:{x}\right)+{f}\left({x}\right)\mathrm{sin}\:{x}}{\left(\mathrm{1}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\right]{dx} \\ $$$$=\int{e}^{{x}} \left[\frac{\mathrm{1}−\mathrm{sin}\:{x}+\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\mathrm{cos}\:{x}}{\left(\mathrm{1}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }\right]{dx} \\ $$$$\Rightarrow\:{f}\left({x}\right)+{f}\left({x}\right)\mathrm{cos}\:{x}+{f}\:'\left({x}\right)\left(\mathrm{1}+\mathrm{cos}\:{x}\right)+{f}\left({x}\right)\mathrm{sin}\:{x} \\ $$$$\:\:\:\:\:\:\:=\:\mathrm{1}−\mathrm{sin}\:{x}+\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\mathrm{cos}\:{x} \\ $$$$\Rightarrow{f}\left({x}\right)\left(\mathrm{1}+\mathrm{cos}\:{x}+\mathrm{sin}\:{x}\right)+{f}\:'\left({x}\right)\left(\mathrm{1}+\mathrm{cos}\:{x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:=\:\mathrm{1}−\mathrm{sin}\:{x}+\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\mathrm{cos}\:{x} \\ $$$$\Rightarrow\:\:{f}\:'\left({x}\right)+\frac{\mathrm{1}+\mathrm{cos}\:{x}+\mathrm{sin}\:{x}}{\mathrm{1}+\mathrm{cos}\:{x}}{f}\left({x}\right)=\mathrm{1}−\mathrm{sin}\:{x} \\ $$$${e}^{{x}} {f}\left({x}\right)=\int{e}^{\int\left(\mathrm{1}+\mathrm{cot}\:\frac{{x}}{\mathrm{2}}\right){dx}} \left(\mathrm{1}−\mathrm{sin}\:{x}\right){dx} \\ $$$$\Rightarrow\:\:{e}^{{x}} {f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\int{e}^{{x}} \left(\mathrm{1}−\mathrm{cos}\:{x}\right)\left(\mathrm{1}−\mathrm{sin}\:{x}\right){dx} \\ $$$$….. \\ $$