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e-x-sin-2-x-dx-




Question Number 33494 by mondodotto@gmail.com last updated on 17/Apr/18
 ∫(e^x /(sin^2 x    ))dx
$$\:\int\frac{\boldsymbol{\mathrm{e}}^{\boldsymbol{{x}}} }{\boldsymbol{\mathrm{sin}}^{\mathrm{2}} \boldsymbol{{x}}\:\:\:\:}\boldsymbol{{dx}} \\ $$
Commented by math khazana by abdo last updated on 18/Apr/18
∫    (e^x /(sin^2 x))dx = ∫  (e^x /((1−cos(2x))/2))dx  = 2 ∫     (e^x /(1−cos(2x)))dx  =_(2x=t)   2 ∫      (e^(t/2) /(1−cost)) (dt/2)
$$\int\:\:\:\:\frac{{e}^{{x}} }{{sin}^{\mathrm{2}} {x}}{dx}\:=\:\int\:\:\frac{{e}^{{x}} }{\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}{dx} \\ $$$$=\:\mathrm{2}\:\int\:\:\:\:\:\frac{{e}^{{x}} }{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{dx} \\ $$$$=_{\mathrm{2}{x}={t}} \:\:\mathrm{2}\:\int\:\:\:\:\:\:\frac{{e}^{\frac{{t}}{\mathrm{2}}} }{\mathrm{1}−{cost}}\:\frac{{dt}}{\mathrm{2}} \\ $$
Commented by math khazana by abdo last updated on 18/Apr/18
∫  (e^x /(sin^2 x))dx  = ∫  e^(t/2)   (Σ_(n=0) ^∞  cos^n t)dt  = Σ_(n=0) ^∞    ∫   e^(t/2)   cos^n t dt  = Σ_(n=0) ^∞   A_n    with  A_n  = ∫  e^(t/2)  cos^n t dt  integral calculable by recurence
$$\int\:\:\frac{{e}^{{x}} }{{sin}^{\mathrm{2}} {x}}{dx}\:\:=\:\int\:\:{e}^{\frac{{t}}{\mathrm{2}}} \:\:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{cos}^{{n}} {t}\right){dt} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\int\:\:\:{e}^{\frac{{t}}{\mathrm{2}}} \:\:{cos}^{{n}} {t}\:{dt}\:\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{A}_{{n}} \:\:\:{with} \\ $$$${A}_{{n}} \:=\:\int\:\:{e}^{\frac{{t}}{\mathrm{2}}} \:{cos}^{{n}} {t}\:{dt}\:\:{integral}\:{calculable}\:{by}\:{recurence} \\ $$

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