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Question Number 133360 by liberty last updated on 21/Feb/21

∫ (√(1+sec x)) dx ?

$$\int\:\sqrt{\mathrm{1}+\mathrm{sec}\:\mathrm{x}}\:\mathrm{dx}\:? \\ $$

Commented by som(math1967) last updated on 21/Feb/21

∫(√((1+cosx)/(cosx)))dx (√2)∫((cos(x/2))/( (√(1−2sin^2 (x/2)))))dx =((√2)/( (√2)))∫((cos(x/2))/( (√(((1/( (√2))))^2 −sin^2 (x/2)))))dx =2∫((cos(x/2))/( 2(√(((1/( (√2))))^2 −sin^2 (x/2)))))dx =2∫((d(sin(x/2)))/( (√(((1/( (√2))))^2 −sin^2 (x/2))))) =2sin^(−1) ((sin(x/2))/(1/( (√2))))+c =2sin^(−1) (√2)sin(x/2)+c

$$\int\sqrt{\frac{\mathrm{1}+{cosx}}{{cosx}}}{dx} \\ $$$$\sqrt{\mathrm{2}}\int\frac{{cos}\frac{{x}}{\mathrm{2}}}{\:\sqrt{\mathrm{1}−\mathrm{2}{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}}{dx} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}\int\frac{{cos}\frac{{x}}{\mathrm{2}}}{\:\sqrt{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}}{dx} \\ $$$$=\mathrm{2}\int\frac{{cos}\frac{{x}}{\mathrm{2}}}{\:\mathrm{2}\sqrt{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}}{dx} \\ $$$$=\mathrm{2}\int\frac{{d}\left({sin}\frac{{x}}{\mathrm{2}}\right)}{\:\sqrt{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} −{sin}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}} \\ $$$$=\mathrm{2sin}^{−\mathrm{1}} \frac{{sin}\frac{{x}}{\mathrm{2}}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}+{c} \\ $$$$=\mathrm{2sin}^{−\mathrm{1}} \sqrt{\mathrm{2}}{sin}\frac{{x}}{\mathrm{2}}+{c} \\ $$$$ \\ $$

Commented by liberty last updated on 22/Feb/21

in other way I = ∫ ((√(1+cos x))/( (√(cos x)))) dx let cos x = z^2 → dx =−((2z)/(sin x)) dz dx = −((2z)/( (√(1−cos^2 x)))) dz=−((2z)/( (√(1−z^4 )))) dz I= ∫ ((√(1+z^2 ))/z) (−((2z)/( (√((1+z^2 )(1−z^2 )))))dz) I=−2∫ (dz/( (√(1−z^2 )))) =−2sin^(−1) (z)+c I=−2sin^(−1) ((√(cos x)) )+ c

$$\mathrm{in}\:\mathrm{other}\:\mathrm{way}\: \\ $$$$\mathrm{I}\:=\:\int\:\frac{\sqrt{\mathrm{1}+\mathrm{cos}\:\mathrm{x}}}{\:\sqrt{\mathrm{cos}\:\mathrm{x}}}\:\mathrm{dx}\: \\ $$$$\mathrm{let}\:\mathrm{cos}\:\mathrm{x}\:=\:\mathrm{z}^{\mathrm{2}} \:\rightarrow\:\mathrm{dx}\:=−\frac{\mathrm{2z}}{\mathrm{sin}\:\mathrm{x}}\:\mathrm{dz} \\ $$$$\:\mathrm{dx}\:=\:−\frac{\mathrm{2z}}{\:\sqrt{\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}}\:\mathrm{dz}=−\frac{\mathrm{2z}}{\:\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{4}} }}\:\mathrm{dz} \\ $$$$\mathrm{I}=\:\int\:\frac{\sqrt{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }}{\mathrm{z}}\:\left(−\frac{\mathrm{2z}}{\:\sqrt{\left(\mathrm{1}+\mathrm{z}^{\mathrm{2}} \right)\left(\mathrm{1}−\mathrm{z}^{\mathrm{2}} \right)}}\mathrm{dz}\right) \\ $$$$\mathrm{I}=−\mathrm{2}\int\:\frac{\mathrm{dz}}{\:\sqrt{\mathrm{1}−\mathrm{z}^{\mathrm{2}} }}\:=−\mathrm{2sin}^{−\mathrm{1}} \left(\mathrm{z}\right)+\mathrm{c} \\ $$$$\mathrm{I}=−\mathrm{2sin}^{−\mathrm{1}} \left(\sqrt{\mathrm{cos}\:\mathrm{x}}\:\right)+\:\mathrm{c}\: \\ $$

Answered by liberty last updated on 21/Feb/21

$$ \\ $$

Commented by liberty last updated on 21/Feb/21

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