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Question Number 122940 by CanovasCamiseros last updated on 21/Nov/20

Commented by CanovasCamiseros last updated on 21/Nov/20

how can i find it?

$$\boldsymbol{{how}}\:\boldsymbol{{can}}\:\boldsymbol{{i}}\:\boldsymbol{{find}}\:\boldsymbol{{it}}? \\ $$

Commented by mr W last updated on 21/Nov/20

go to Q73012

$${go}\:{to}\:{Q}\mathrm{73012} \\ $$

Answered by TANMAY PANACEA last updated on 21/Nov/20

t^2 =tanx→2t(dt/dx)=sec^2 x  dx=((2t)/(1+t^4 ))dt  ∫((t×2tdt)/(1+t^4 ))  ∫((2dt)/(t^2 +(1/t^2 )))  ∫((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))  use formula∫(dx/(x^2 −a^2 )) &((∫dx)/(x^2 +a^2 ))  (1/(2(√2)))ln(((t+(1/t))−(√2))/((t+(1/t)+(√2))))+(1/( (√2)))tan^(−1) (((t−(1/t))/( (√2))))  puy   t=(√(tanx))

$${t}^{\mathrm{2}} ={tanx}\rightarrow\mathrm{2}{t}\frac{{dt}}{{dx}}={sec}^{\mathrm{2}} {x} \\ $$$${dx}=\frac{\mathrm{2}{t}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$\int\frac{{t}×\mathrm{2}{tdt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\int\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }} \\ $$$$\int\frac{\mathrm{1}−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }+\mathrm{1}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{\mathrm{1}}{{t}^{\mathrm{2}} }}{dt} \\ $$$$\int\frac{{d}\left({t}+\frac{\mathrm{1}}{{t}}\right)}{\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} −\mathrm{2}}+\int\frac{{d}\left({t}−\frac{\mathrm{1}}{{t}}\right)}{\left({t}−\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}} \\ $$$${use}\:{formula}\int\frac{{dx}}{{x}^{\mathrm{2}} −{a}^{\mathrm{2}} }\:\&\frac{\int{dx}}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\left(\frac{\left.{t}+\frac{\mathrm{1}}{{t}}\right)−\sqrt{\mathrm{2}}}{\left({t}+\frac{\mathrm{1}}{{t}}+\sqrt{\mathrm{2}}\right.}\right)+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{t}}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$${puy}\:\:\:{t}=\sqrt{{tanx}}\: \\ $$$$ \\ $$$$ \\ $$

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