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Question Number 131849 by mnjuly1970 last updated on 09/Feb/21

∗∗∗ calculus (I) ∗∗∗ please evaluate:: φ=∫(dx/(sin(2x)ln(tan(x)))) Trinity College Cambridge ....1897...

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\ast\ast\ast\:\:\:{calculus}\:\left({I}\right)\:\ast\ast\ast \\ $$$$\:\:\:{please}\:\:{evaluate}:: \\ $$$$\:\:\:\:\:\:\:\:\phi=\int\frac{{dx}}{{sin}\left(\mathrm{2}{x}\right){ln}\left({tan}\left({x}\right)\right)} \\ $$$$\:\:\:\:\:\:{Trinity}\:{College} \\ $$$$\:\:\:\:\:\:\:{Cambridge}\:....\mathrm{1897}... \\ $$

Answered by mindispower last updated on 09/Feb/21

∅=∫(dx/(2sin(x)cos(x)ln(tg(x)))) =∫((dx/(cos^2 (x)))/(2tg(x)ln(tg(x)))) let u=tg(x) ⇔∫(du/(2u,ln(u)))=(1/2)∫((dln(u))/(ln(u)))=ln(ln(u))+c we get ln(ln(tg(x)))+c

$$\emptyset=\int\frac{{dx}}{\mathrm{2}{sin}\left({x}\right){cos}\left({x}\right){ln}\left({tg}\left({x}\right)\right)} \\ $$$$=\int\frac{\frac{{dx}}{{cos}^{\mathrm{2}} \left({x}\right)}}{\mathrm{2}{tg}\left({x}\right){ln}\left({tg}\left({x}\right)\right)}\:\:{let}\:{u}={tg}\left({x}\right) \\ $$$$\Leftrightarrow\int\frac{{du}}{\mathrm{2}{u},{ln}\left({u}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dln}\left({u}\right)}{{ln}\left({u}\right)}={ln}\left({ln}\left({u}\right)\right)+{c} \\ $$$${we}\:{get}\:{ln}\left({ln}\left({tg}\left({x}\right)\right)\right)+{c} \\ $$

Commented by mnjuly1970 last updated on 09/Feb/21

thanks alot mr mindspower..

$${thanks}\:{alot}\:{mr}\:{mindspower}.. \\ $$

Answered by Dwaipayan Shikari last updated on 09/Feb/21

∫(dx/(2sinx cosx log(tanx))) tanx=t =∫(dt/(2t log(t)))=(1/2)log(log(t))=(1/2)log(log(tanx))+C

$$\int\frac{{dx}}{\mathrm{2}{sinx}\:{cosx}\:{log}\left({tanx}\right)}\:\:\:\:\:{tanx}={t} \\ $$$$=\int\frac{{dt}}{\mathrm{2}{t}\:{log}\left({t}\right)}=\frac{\mathrm{1}}{\mathrm{2}}{log}\left({log}\left({t}\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}{log}\left({log}\left({tanx}\right)\right)+{C} \\ $$

Commented by mnjuly1970 last updated on 09/Feb/21

mercey mr dwaipayan...

$${mercey}\:\:{mr}\:{dwaipayan}... \\ $$

Answered by mnjuly1970 last updated on 09/Feb/21

φ=(1/2)∫ ((tan(x)+cot(x))/(ln(tan(x))))dx =(1/2)∫((1+tan^2 (x))/(tan(x)ln(tan(x))))dx =^(tan(x)=u) (1/2)∫(du/(uln(u)))=(1/2)∫((1/u)/(ln(u)))du =(1/2)ln(ln(tan(x))+C

$$\:\:\:\phi=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{{tan}\left({x}\right)+{cot}\left({x}\right)}{{ln}\left({tan}\left({x}\right)\right)}{dx} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}+{tan}^{\mathrm{2}} \left({x}\right)}{{tan}\left({x}\right){ln}\left({tan}\left({x}\right)\right)}{dx} \\ $$$$\:\:\:\:\overset{{tan}\left({x}\right)={u}} {=}\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{{uln}\left({u}\right)}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\frac{\mathrm{1}}{{u}}}{{ln}\left({u}\right)}{du} \\ $$$$\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({ln}\left({tan}\left({x}\right)\right)+{C}\right. \\ $$

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