Evaluate-tan-d- – Tinku Tara

Question Number 119 by novrya last updated on 25/Jan/15

$${Evaluate}\:\int\sqrt{{tan}\:\theta}\:{d}\theta \\ $$

Answered by rajabhay last updated on 06/Dec/14

tan θ=t^2 sec^2 θ dθ=2t dt, sec^2 θ=1+t^4 dθ=((2t dt)/(1+t^4 )) ∫(√(tanθ )) dθ=∫((2t^2 )/(1+t^4 )) dt =∫ ((t^2 +1)/(1+t^4 ))dt+∫ ((t^2 −1)/(1+t^4 ))dt =∫ ((1+1/t^2 )/(t^2 +1/t^2 )) dt+∫ ((1−1/t^2 )/(t^2 +1/t^2 )) =∫ ((1+1/t^2 )/(t^2 +1/t^2 −2+2)) dt+∫ ((1−1/t^2 )/(t^2 +1/t^2 +2−2)) =∫ ((1+1/t^2 )/((t−1/t)^2 +2)) dt+∫ ((1−1/t^2 )/((t+1/t)^2 −2)) t−1/t=u in 1st intergral and t+1/t=v in 2nd (1+1/t^2 )dt=du, (1−1/t^2 )dt=dv So the integral now equals ∫(du/(u^2 +2)) +∫ (dv/(v^2 −2)) =(1/( (√2)))tan^(−1) (u/( (√2)))+(1/(2(√2)))ln ∣((v−(√2))/(v+(√2)))∣+C =(1/( (√2)))tan^(−1) (1/( (√2)))(t−(1/t))+(1/(2(√2)))ln ∣((t+1/t−(√2))/(t+1/t+(√2)))∣+C =(1/( (√2)))tan^(−1) (1/( (√2)))(((t^2 −1)/t))+(1/(2(√2)))ln ∣((t^2 +1−t(√2))/(t^2 +1+t(√2)))∣+C =(1/( (√2)))tan^(−1) (((tan θ−1)/( (√(2tan θ)))))+(1/(2(√2)))ln ∣((tan θ+1−(√(2tanθ)))/(tan θ+1+(√(2tanθ))))∣+C

$$\mathrm{tan}\:\theta={t}^{\mathrm{2}} \\ $$$$\mathrm{sec}^{\mathrm{2}} \theta\:{d}\theta=\mathrm{2}{t}\:{dt},\:\mathrm{sec}^{\mathrm{2}} \theta=\mathrm{1}+{t}^{\mathrm{4}} \\ $$$${d}\theta=\frac{\mathrm{2}{t}\:{dt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\int\sqrt{\mathrm{tan}\theta\:}\:{d}\theta=\int\frac{\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{4}} }\:{dt} \\ $$$$=\int\:\frac{{t}^{\mathrm{2}} +\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt}+\int\:\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}+{t}^{\mathrm{4}} }{dt} \\ $$$$=\int\:\frac{\mathrm{1}+\mathrm{1}/{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}/{t}^{\mathrm{2}} }\:{dt}+\int\:\frac{\mathrm{1}−\mathrm{1}/{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}/{t}^{\mathrm{2}} } \\ $$$$=\int\:\frac{\mathrm{1}+\mathrm{1}/{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}/{t}^{\mathrm{2}} −\mathrm{2}+\mathrm{2}}\:{dt}+\int\:\frac{\mathrm{1}−\mathrm{1}/{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} +\mathrm{1}/{t}^{\mathrm{2}} +\mathrm{2}−\mathrm{2}} \\ $$$$=\int\:\frac{\mathrm{1}+\mathrm{1}/{t}^{\mathrm{2}} }{\left({t}−\mathrm{1}/{t}\right)^{\mathrm{2}} +\mathrm{2}}\:{dt}+\int\:\frac{\mathrm{1}−\mathrm{1}/{t}^{\mathrm{2}} }{\left({t}+\mathrm{1}/{t}\right)^{\mathrm{2}} −\mathrm{2}} \\ $$$${t}−\mathrm{1}/{t}={u}\:\mathrm{in}\:\mathrm{1st}\:\mathrm{intergral}\:\mathrm{and}\:{t}+\mathrm{1}/{t}={v}\:\mathrm{in}\:\mathrm{2nd} \\ $$$$\left(\mathrm{1}+\mathrm{1}/{t}^{\mathrm{2}} \right){dt}={du},\:\left(\mathrm{1}−\mathrm{1}/{t}^{\mathrm{2}} \right){dt}={dv} \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{now}\:\mathrm{equals} \\ $$$$\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{2}}\:+\int\:\frac{{dv}}{{v}^{\mathrm{2}} −\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \frac{{u}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\:\mid\frac{{v}−\sqrt{\mathrm{2}}}{{v}+\sqrt{\mathrm{2}}}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left({t}−\frac{\mathrm{1}}{{t}}\right)+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\:\mid\frac{{t}+\mathrm{1}/{t}−\sqrt{\mathrm{2}}}{{t}+\mathrm{1}/{t}+\sqrt{\mathrm{2}}}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left(\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}}\right)+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\:\mid\frac{{t}^{\mathrm{2}} +\mathrm{1}−{t}\sqrt{\mathrm{2}}}{{t}^{\mathrm{2}} +\mathrm{1}+{t}\sqrt{\mathrm{2}}}\mid+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{tan}\:\theta−\mathrm{1}}{\:\sqrt{\mathrm{2tan}\:\theta}}\right)+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\:\mid\frac{\mathrm{tan}\:\theta+\mathrm{1}−\sqrt{\mathrm{2tan}\theta}}{\mathrm{tan}\:\theta+\mathrm{1}+\sqrt{\mathrm{2tan}\theta}}\mid+{C} \\ $$

Commented by malwaan last updated on 07/Sep/16

$${greet} \\ $$

Commented by rajabhay last updated on 25/Sep/16

$$\mathrm{Thanks} \\ $$

Commented by peter frank last updated on 05/Jan/22

$$\mathrm{thanks} \\ $$

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