tan-x-dx-

Question Number 78575 by TawaTawa last updated on 18/Jan/20

$$\int\:\sqrt{\mathrm{tan}\:\mathrm{x}}\:\:\mathrm{dx} \\ $$

Answered by peter frank last updated on 18/Jan/20

u=(√(tan x)) dx=((2udu)/(1+u^4 )) ∫((2u^2 )/(1+u^4 ))du ∫((u^2 +u^2 +1−1)/(1+u^4 ))du ∫((u^2 −1)/(1+u^4 ))+∫((u^2 +1)/(1+u^4 ))du I=I_1 +I_2 ∫((1−(1/u^2 ))/(u^2 +(1/u^2 )))du=I_1 .....(i) ∫((1+(1/u^2 ))/(u^2 +(1/u^2 )))du=I_2 .....(ii) from (i) ∫((1−(1/u^2 ))/(u^2 +(1/u^2 )))du=I_1 ∫((1−(1/u^2 ))/(u^2 +(1/u^2 )))du=I_1 ∫((1−(1/u^2 ))/(u^2 +(1/u^2 )+2−2))du=I_1 let P=(u+(1/u)) ∫((1−(1/u^2 ))/(p^2 −2)).(dp/(1−(1/u^2 )))=I_1 ∫(dp/(p^2 −2))=I_1 (1/(2(√2)))ln(((p−(√2))/(p+(√p))))......(iii) from ∫((1+(1/u^2 ))/(u^2 +(1/u^2 )−2+2))du=I_2 .....(ii) let w=(u−(1/u)) ∫(dp/(w^2 +2))=I_2 (1/( (√2)))tan^(−1) (w/( (√2))) (1/(2(√2)))ln(((p−(√2))/(p+(√p))))+(1/( (√2)))tan^(−1) (w/( (√2)))

$${u}=\sqrt{\mathrm{tan}\:{x}} \\ $$$${dx}=\frac{\mathrm{2}{udu}}{\mathrm{1}+{u}^{\mathrm{4}} } \\ $$$$\int\frac{\mathrm{2}{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{4}} }{du} \\ $$$$\int\frac{{u}^{\mathrm{2}} +{u}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{4}} }{du} \\ $$$$\int\frac{{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{4}} }+\int\frac{{u}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{4}} }{du} \\ $$$${I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$$\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{du}={I}_{\mathrm{1}} …..\left({i}\right) \\ $$$$\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{du}={I}_{\mathrm{2}} …..\left({ii}\right) \\ $$$${from}\:\left({i}\right) \\ $$$$\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{du}={I}_{\mathrm{1}} \\ $$$$\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{du}={I}_{\mathrm{1}} \\ $$$$\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }+\mathrm{2}−\mathrm{2}}{du}={I}_{\mathrm{1}} \\ $$$${let}\:{P}=\left({u}+\frac{\mathrm{1}}{{u}}\right) \\ $$$$\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{p}^{\mathrm{2}} −\mathrm{2}}.\frac{{dp}}{\mathrm{1}−\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}={I}_{\mathrm{1}} \\ $$$$\int\frac{{dp}}{{p}^{\mathrm{2}} −\mathrm{2}}={I}_{\mathrm{1}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{{p}−\sqrt{\mathrm{2}}}{{p}+\sqrt{{p}}}\right)……\left({iii}\right) \\ $$$${from} \\ $$$$\int\frac{\mathrm{1}+\frac{\mathrm{1}}{{u}^{\mathrm{2}} }}{{u}^{\mathrm{2}} +\frac{\mathrm{1}}{{u}^{\mathrm{2}} }−\mathrm{2}+\mathrm{2}}{du}={I}_{\mathrm{2}} …..\left({ii}\right) \\ $$$${let}\:{w}=\left({u}−\frac{\mathrm{1}}{{u}}\right) \\ $$$$\int\frac{{dp}}{{w}^{\mathrm{2}} +\mathrm{2}}={I}_{\mathrm{2}} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \frac{{w}}{\:\sqrt{\mathrm{2}}} \\ $$$$ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{{p}−\sqrt{\mathrm{2}}}{{p}+\sqrt{{p}}}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \frac{{w}}{\:\sqrt{\mathrm{2}}} \\ $$$$ \\ $$

Commented by TawaTawa last updated on 18/Jan/20

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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