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Question Number 155883 by john_santu last updated on 05/Oct/21

∫ ((tan^2 x)/(1−tan^2 x)) dx=?

$$\int\:\frac{\mathrm{tan}\:^{\mathrm{2}} {x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}}\:{dx}=? \\ $$

Commented by aliyn last updated on 05/Oct/21

I = ∫ ((tan^2 x)/(1 − tan^2 x)) dx Solution: I = ∫ ((sec^2 x − 1)/(sec^2 x ( cos^2 x − sin^2 x ))) dx I = ∫ ((sec^2 x − 1)/(sec^2 x ( cos2x))) dx = ∫ (dx/(cos2x)) − ∫ ((cos^2 x)/(cos2x)) dx I = ∫ sec2x dx − (1/2) ∫ ((( 1+ cos2x))/(cos2x)) dx I = (1/2) ∫ sec2x × ((sec2x + tan2x)/(sec2x + tan2x)) dx − (1/2) ∫ (1) dx I = (1/4) ln ∣ sec2x + tan2x ∣ − (1/2) x + C □ M

$$\boldsymbol{{I}}\:=\:\int\:\frac{\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{x}}}{\mathrm{1}\:−\:\boldsymbol{{tan}}^{\mathrm{2}} \boldsymbol{{x}}}\:\boldsymbol{{dx}} \\ $$$$ \\ $$$$\boldsymbol{{Solution}}: \\ $$$$ \\ $$$$\boldsymbol{{I}}\:=\:\int\:\frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}\:−\:\mathrm{1}}{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}\:\left(\:\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}\:−\:\boldsymbol{{sin}}^{\mathrm{2}} \boldsymbol{{x}}\:\right)}\:\boldsymbol{{dx}}\: \\ $$$$ \\ $$$$\boldsymbol{{I}}\:=\:\int\:\frac{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}\:−\:\mathrm{1}}{\boldsymbol{{sec}}^{\mathrm{2}} \boldsymbol{{x}}\:\left(\:\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}\right)}\:\boldsymbol{{dx}}\:=\:\int\:\frac{\boldsymbol{{dx}}}{\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}}\:−\:\int\:\frac{\boldsymbol{{cos}}^{\mathrm{2}} \boldsymbol{{x}}}{\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}}\:\boldsymbol{{dx}} \\ $$$$ \\ $$$$\boldsymbol{{I}}\:=\:\int\:\boldsymbol{{sec}}\mathrm{2}\boldsymbol{{x}}\:\boldsymbol{{dx}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\frac{\left(\:\mathrm{1}+\:\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}\right)}{\boldsymbol{{cos}}\mathrm{2}\boldsymbol{{x}}}\:\boldsymbol{{dx}}\: \\ $$$$ \\ $$$$\boldsymbol{{I}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\boldsymbol{{sec}}\mathrm{2}\boldsymbol{{x}}\:×\:\frac{\boldsymbol{{sec}}\mathrm{2}\boldsymbol{{x}}\:+\:\boldsymbol{{tan}}\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{sec}}\mathrm{2}\boldsymbol{{x}}\:+\:\boldsymbol{{tan}}\mathrm{2}\boldsymbol{{x}}}\:\boldsymbol{{dx}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\left(\mathrm{1}\right)\:\boldsymbol{{dx}} \\ $$$$ \\ $$$$\boldsymbol{{I}}\:=\:\frac{\mathrm{1}}{\mathrm{4}}\:\boldsymbol{{ln}}\:\mid\:\boldsymbol{{sec}}\mathrm{2}\boldsymbol{{x}}\:+\:\boldsymbol{{tan}}\mathrm{2}\boldsymbol{{x}}\:\mid\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\boldsymbol{{x}}\:+\:\boldsymbol{{C}} \\ $$$$ \\ $$$$\square\:\boldsymbol{{M}}\: \\ $$

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