Question-197396

Question Number 197396 by sonukgindia last updated on 16/Sep/23

Commented by Frix last updated on 16/Sep/23

We had this several times before. t=(√(tan x)) ⇒ 2∫(t^2 /(t^4 +1))dt which can be solved by decomposing

$$\mathrm{We}\:\mathrm{had}\:\mathrm{this}\:\mathrm{several}\:\mathrm{times}\:\mathrm{before}. \\ $$$${t}=\sqrt{\mathrm{tan}\:{x}}\:\Rightarrow\:\mathrm{2}\int\frac{{t}^{\mathrm{2}} }{{t}^{\mathrm{4}} +\mathrm{1}}{dt} \\ $$$$\mathrm{which}\:\mathrm{can}\:\mathrm{be}\:\mathrm{solved}\:\mathrm{by}\:\mathrm{decomposing} \\ $$

Answered by universe last updated on 16/Sep/23

2I = ∫((√(tan x)) + (√(cot x)) )_(I_1 ) dx + ∫((√(tan x)) − (√(cot x )) )dx_(I_2 ) I_1 = (√2)∫((sin x + cos x)/( (√(2sin xcos x))))dx I_1 = (√2)∫(((sin x +cos x))/( (√(1−(sin x − cos x)^2 ))))dx let sin x − cos x = t ⇒ (cos x +sin x)dx=dt I_1 = (√2)∫(dt/( (√(1−t^2 )))) = (√2)sin^(−1) t + C_1 I_(1 ) = (√2) sin^(−1) (sin x − cos x) + C_1 now I_(2 ) = −(√2)∫(((−sin x +cos x))/( (√((sin x + cos x)^2 −1))))dx let sin x + cos x = y (cos x−sin x)dx= dy I_2 = −(√2) ∫(dy/( (√(y^2 −1)))) I_2 = −(√2) cosh^(−1) y + C_2 I_(2 ) = −(√2) cosh^(−1) (sin x+cos x) + C_2 2I = I_1 + I_2 I =(1/( (√2)))[sin^(−1) (sinx−cosx)−cosh^(−1) (sinx+cosx)+ C

$$\:\mathrm{2}{I}\:\:=\:\int\underset{{I}_{\mathrm{1}} } {\underbrace{\left(\sqrt{\mathrm{tan}\:{x}}\:+\:\sqrt{\mathrm{cot}\:{x}}\:\right)}{dx}}\:+\underset{{I}_{\mathrm{2}} } {\underbrace{\:\int\left(\sqrt{\mathrm{tan}\:{x}}\:−\:\sqrt{\mathrm{cot}\:{x}\:}\:\right){dx}}} \\ $$$${I}_{\mathrm{1}} \:\:=\:\:\sqrt{\mathrm{2}}\int\frac{\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{x}}{\:\sqrt{\mathrm{2sin}\:{x}\mathrm{cos}\:{x}}}{dx} \\ $$$${I}_{\mathrm{1}} \:=\:\:\sqrt{\mathrm{2}}\int\frac{\left(\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)}{\:\sqrt{\mathrm{1}−\left(\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\right)^{\mathrm{2}} }}{dx} \\ $$$${let}\: \\ $$$$\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\:=\:{t}\:\:\Rightarrow\:\left(\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\right){dx}={dt} \\ $$$$\:\:{I}_{\mathrm{1}} \:=\:\sqrt{\mathrm{2}}\int\frac{{dt}}{\:\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:=\:\sqrt{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} {t}\:+\:{C}_{\mathrm{1}} \\ $$$$\:\:{I}_{\mathrm{1}\:} \:=\:\:\sqrt{\mathrm{2}}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\right)\:+\:{C}_{\mathrm{1}} \\ $$$${now}\: \\ $$$${I}_{\mathrm{2}\:} =\:−\sqrt{\mathrm{2}}\int\frac{\left(−\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right)}{\:\sqrt{\left(\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{x}\right)^{\mathrm{2}} −\mathrm{1}}}{dx} \\ $$$$\:\:{let}\:\mathrm{sin}\:{x}\:+\:\mathrm{cos}\:{x}\:=\:{y}\: \\ $$$$\:\left(\mathrm{cos}\:{x}−\mathrm{sin}\:{x}\right){dx}=\:{dy} \\ $$$${I}_{\mathrm{2}} \:=\:−\sqrt{\mathrm{2}}\:\int\frac{{dy}}{\:\sqrt{{y}^{\mathrm{2}} −\mathrm{1}}} \\ $$$$\:{I}_{\mathrm{2}} \:\:=\:\:−\sqrt{\mathrm{2}}\:\mathrm{cosh}^{−\mathrm{1}} {y}\:+\:{C}_{\mathrm{2}} \\ $$$$\:\:{I}_{\mathrm{2}\:} \:\:=\:\:−\sqrt{\mathrm{2}}\:\mathrm{cosh}^{−\mathrm{1}} \left(\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)\:\:+\:{C}_{\mathrm{2}} \: \\ $$$$\mathrm{2}{I}\:\:=\:\:{I}_{\mathrm{1}} \:+\:{I}_{\mathrm{2}} \\ $$$${I}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\left[\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{sin}{x}−\mathrm{cos}{x}\right)−\mathrm{cosh}^{−\mathrm{1}} \left(\mathrm{sin}{x}+\mathrm{cos}{x}\right)+\:{C}\right. \\ $$

Commented by MM42 last updated on 16/Sep/23

$$\:\cancel{\lesseqgtr} \\ $$

Commented by universe last updated on 16/Sep/23

$$\:\underline{\underbrace{\lesseqgtr}} \\ $$

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