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Question Number 51122 by behi83417@gmail.com last updated on 24/Dec/18

∫ (dx/(tgx−(√(tgx))))=?

$$\int\:\:\:\:\frac{\boldsymbol{\mathrm{dx}}}{\boldsymbol{\mathrm{tgx}}−\sqrt{\boldsymbol{\mathrm{tgx}}}}=? \\ $$

Commented by MJS last updated on 24/Dec/18

long way... start with t=(√(tan x)) → dx=((2t)/(t^4 +1))dt ⇒ 2∫(dt/((t+1)(t^4 +1))) and continue with decomposing...

$$\mathrm{long}\:\mathrm{way}... \\ $$$$\mathrm{start}\:\mathrm{with} \\ $$$${t}=\sqrt{\mathrm{tan}\:{x}}\:\rightarrow\:{dx}=\frac{\mathrm{2}{t}}{{t}^{\mathrm{4}} +\mathrm{1}}{dt} \\ $$$$\Rightarrow\:\mathrm{2}\int\frac{{dt}}{\left({t}+\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{1}\right)}\:\mathrm{and}\:\mathrm{continue}\:\mathrm{with}\:\mathrm{decomposing}... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18

∫(dx/(tanx−(√(tanx)))) t^2 =tanx 2tdt=sec^2 xdx ((2t)/(1+t^4 ))dt=dx ∫(dx/((√(tanx)) ((√(tanx)) −1))) ∫(((√(tanx)) −((√(tanx)) −1))/((√(tanx)) ((√(tanx)) −1)))dx ∫(dx/((√(tanx)) −1))−∫(dx/(√(tanx))) I_1 −I_2 I_2 =∫((2t)/(1+t^4 ))×(1/t)dt =∫((2dt)/(1+t^4 )) ∫((2/t^2 )/(t^2 +(1/t^2 )))dt ∫(((1+(1/t^2 ))−(1−(1/t^2 )))/((t^2 +(1/t^2 ))))dt ∫((d(t−(1/t)))/((t−(1/t))^2 +2))−∫((d(t+(1/t)))/((t+(1/t))^2 −2)) (1/(√2))tan^(−1) (((t−(1/t))/(√2)))−(1/(2(√2)))ln(((t+(1/t)−(√2))/(t+(1/t)+(√2))))+c_1 (1/(√2))tan^(−1) ((((√(tanx)) −(1/(√(tanx))))/(√2)))−(1/(2(√2)))ln((((√(tanx)) +(1/(√(tanx)))−(√2))/((√(tanx)) +(1/((√(tanx)) ))+(√2))))+c_1 I_1 =∫(dx/((√(tanx)) −1)) =∫((2tdt)/(1+t^4 ))×(1/(t−1)) 2∫((t−1+1)/((t^4 +1)(t−1)))dt 2∫(dt/(t^4 +1))+2∫(dt/((t−1)(t^4 +1)))←SOLVE IT... I_2 +2I_3

$$\int\frac{{dx}}{{tanx}−\sqrt{{tanx}}} \\ $$$${t}^{\mathrm{2}} ={tanx}\:\:\mathrm{2}{tdt}={sec}^{\mathrm{2}} {xdx} \\ $$$$\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}={dx} \\ $$$$\int\frac{{dx}}{\sqrt{{tanx}}\:\left(\sqrt{{tanx}}\:−\mathrm{1}\right)} \\ $$$$\int\frac{\sqrt{{tanx}}\:−\left(\sqrt{{tanx}}\:−\mathrm{1}\right)}{\sqrt{{tanx}}\:\left(\sqrt{{tanx}}\:−\mathrm{1}\right)}{dx} \\ $$$$\int\frac{{dx}}{\sqrt{{tanx}}\:−\mathrm{1}}−\int\frac{{dx}}{\sqrt{{tanx}}} \\ $$$${I}_{\mathrm{1}} −{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{2}} =\int\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{4}} }×\frac{\mathrm{1}}{{t}}{dt} \\ $$$$=\int\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\int\frac{\frac{\mathrm{2}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$\int\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)−\left(\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}{\left({t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)}{dt} \\ $$$$\int\frac{{d}\left({t}−\frac{\mathrm{1}}{{t}}\right)}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}}−\int\frac{{d}\left({t}+\frac{\mathrm{1}}{{t}}\right)}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{{t}+\frac{\mathrm{1}}{{t}}−\sqrt{\mathrm{2}}}{{t}+\frac{\mathrm{1}}{{t}}+\sqrt{\mathrm{2}}}\right)+{c}_{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{tanx}}\:−\frac{\mathrm{1}}{\sqrt{{tanx}}}}{\sqrt{\mathrm{2}}}\right)−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{\sqrt{{tanx}}\:+\frac{\mathrm{1}}{\sqrt{{tanx}}}−\sqrt{\mathrm{2}}}{\sqrt{{tanx}}\:+\frac{\mathrm{1}}{\sqrt{{tanx}}\:}+\sqrt{\mathrm{2}}}\right)+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{1}} =\int\frac{{dx}}{\sqrt{{tanx}}\:−\mathrm{1}} \\ $$$$=\int\frac{\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} }×\frac{\mathrm{1}}{{t}−\mathrm{1}} \\ $$$$\mathrm{2}\int\frac{{t}−\mathrm{1}+\mathrm{1}}{\left({t}^{\mathrm{4}} +\mathrm{1}\right)\left({t}−\mathrm{1}\right)}{dt} \\ $$$$\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{4}} +\mathrm{1}}+\mathrm{2}\int\frac{{dt}}{\left({t}−\mathrm{1}\right)\left({t}^{\mathrm{4}} +\mathrm{1}\right)}\leftarrow\boldsymbol{{S}}{OLVE}\:{IT}... \\ $$$${I}_{\mathrm{2}} +\mathrm{2}{I}_{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by behi83417@gmail.com last updated on 25/Dec/18

thank you sir tanmay.

$${thank}\:{you}\:{sir}\:{tanmay}. \\ $$

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