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Question Number 907 by Yugi last updated on 19/Apr/15
Determine the results of the following.                   I_1 =∫(√((a+e^x )/(a−e^x )))dx                  I_2 =∫(((tanx)∣tanx∣)/(1−tan^2 x))dx
$${Determine}\:{the}\:{results}\:{of}\:{the}\:{following}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}_{\mathrm{1}} =\int\sqrt{\frac{{a}+{e}^{{x}} }{{a}−{e}^{{x}} }}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{I}_{\mathrm{2}} =\int\frac{\left({tanx}\right)\mid{tanx}\mid}{\mathrm{1}−{tan}^{\mathrm{2}} {x}}{dx} \\ $$
Commented by prakash jain last updated on 21/Apr/15
I_1   e^x =acos t  e^x dx=−asin t dt  dx=−tan t dt  ∫−tan t (1/(tan (t/2))) dt  =−∫(2/(1−tan^2 t/2)) dt=−∫(sec t+1)dt  =−t−ln (sec t+tan t)+C  where t=cos^(−1) (e^x /a)
$${I}_{\mathrm{1}} \\ $$$${e}^{{x}} ={a}\mathrm{cos}\:{t} \\ $$$${e}^{{x}} {dx}=−{a}\mathrm{sin}\:{t}\:{dt} \\ $$$${dx}=−\mathrm{tan}\:{t}\:{dt} \\ $$$$\int−\mathrm{tan}\:{t}\:\frac{\mathrm{1}}{\mathrm{tan}\:\frac{{t}}{\mathrm{2}}}\:{dt} \\ $$$$=−\int\frac{\mathrm{2}}{\mathrm{1}−\mathrm{tan}^{\mathrm{2}} {t}/\mathrm{2}}\:{dt}=−\int\left(\mathrm{sec}\:{t}+\mathrm{1}\right){dt} \\ $$$$=−{t}−\mathrm{ln}\:\left(\mathrm{sec}\:{t}+\mathrm{tan}\:{t}\right)+{C} \\ $$$$\mathrm{where}\:{t}=\mathrm{cos}^{−\mathrm{1}} \frac{{e}^{{x}} }{{a}} \\ $$

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