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e-x-1-sin-x-1-cos-x-dx-




Question Number 44994 by manish09@gmail.com last updated on 07/Oct/18
∫e^x (((1+sin x)/(1−cos x)))dx = ?
$$\int\mathrm{e}^{\mathrm{x}} \left(\frac{\mathrm{1}+\mathrm{sin}\:\mathrm{x}}{\mathrm{1}−\mathrm{cos}\:\mathrm{x}}\right)\mathrm{dx}\:=\:? \\ $$
Answered by arvinddayama01@gmail.com last updated on 07/Oct/18
i think problem should be:−        ∫e^x (((1−sin x)/(1−cosx)))dx  solved:−       =∫e^x {((1−2sin(x/2)cos(x/2))/(2sin^2 (x/2)))}dx       =∫e^x {(1/2)cosec^2 (x/2)−cot(x/2)}dx       ∵∫e^x {f(x)+f^′ (x)}dx=e^x f(x)+C    so:−         =e^x (−cot(x/2))+C         =−e^x cot(x/2)+C
$$\mathrm{i}\:\mathrm{think}\:\mathrm{problem}\:\mathrm{should}\:\mathrm{be}:− \\ $$$$\:\:\:\:\:\:\int\mathrm{e}^{\mathrm{x}} \left(\frac{\mathrm{1}−\mathrm{sin}\:\mathrm{x}}{\mathrm{1}−\mathrm{cosx}}\right)\mathrm{dx} \\ $$$$\mathrm{solved}:− \\ $$$$\:\:\:\:\:=\int\mathrm{e}^{\mathrm{x}} \left\{\frac{\mathrm{1}−\mathrm{2sin}\frac{\mathrm{x}}{\mathrm{2}}\mathrm{cos}\frac{\mathrm{x}}{\mathrm{2}}}{\mathrm{2sin}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}}\right\}\mathrm{dx} \\ $$$$\:\:\:\:\:=\int\mathrm{e}^{\mathrm{x}} \left\{\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cosec}^{\mathrm{2}} \frac{\mathrm{x}}{\mathrm{2}}−\mathrm{cot}\frac{\mathrm{x}}{\mathrm{2}}\right\}\mathrm{dx} \\ $$$$\:\:\:\:\:\because\int\mathrm{e}^{\mathrm{x}} \left\{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}^{'} \left(\mathrm{x}\right)\right\}\mathrm{dx}=\mathrm{e}^{\mathrm{x}} \mathrm{f}\left(\mathrm{x}\right)+\mathrm{C} \\ $$$$\:\:\mathrm{so}:− \\ $$$$\:\:\:\:\:\:\:=\mathrm{e}^{\mathrm{x}} \left(−\mathrm{cot}\frac{\mathrm{x}}{\mathrm{2}}\right)+\mathrm{C} \\ $$$$\:\:\:\:\:\:\:=−\mathrm{e}^{\mathrm{x}} \mathrm{cot}\frac{\mathrm{x}}{\mathrm{2}}+\mathrm{C} \\ $$

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