tanxdx\int { \sqrt { \tan { x } } } \, dx

tanx dx=u=lettanx2u2u4+1du=2u2+1u2du=1+1u2(u1u)2+2du+11u2(u+1u)22du=v=letu1uw=letu+1u1v2+2dv+1w22dw=12tan1v2+c112tanh1w2+c2=22tan1(u1u)2222tanh1(u+1u)22+c=22(tan1(tanxcotx)22tanh1(tanx+cotx)22)+c\begin{aligned} \int { \sqrt { \tan { x } } } \ dx & \overset { u\overset { let }{ = } \sqrt { \tan { x } } }{ = } \int { \frac { 2u^2 }{ { u }^{ 4 }+1 } du } \\ \quad & = \int { \frac { 2 }{ { u }^{ 2 }+\frac { 1 }{ { u }^{ 2 } } }du } \\ \quad & = \int { \frac { 1+\frac { 1 }{ { u }^{ 2 } } }{ { (u-\frac { 1 }{ u } ) }^{ 2 }+2 } du } +\int { \frac { 1-\frac { 1 }{ { u }^{ 2 } } }{ { (u+\frac { 1 }{ u } ) }^{ 2 }-2 } du } \\ \quad & \overset { \begin{matrix} v\overset { let }{ = } u-\frac { 1 }{ u } \\ w\overset { let }{ = } u+\frac { 1 }{ u } \end{matrix} }{ = } \int { \frac { 1 }{ { v }^{ 2 }+2 } dv } +\int { \frac { 1 }{ { w }^{ 2 }-2 } dw } \\ \quad & = \frac { 1 }{ \sqrt { 2 } } \tan ^{ -1 }{ \frac { v }{ \sqrt { 2 } } +{ c }_{ 1 } } -\frac { 1 }{ \sqrt { 2 } } \tanh ^{ -1 }{ \frac { w }{ \sqrt { 2 } } } +{ c }_{ 2 } \\ \quad & = \frac { \sqrt { 2 } }{ 2 } \tan ^{ -1 }{ \frac { (u-\frac { 1 }{ u } )\sqrt { 2 } }{ 2 } } -\frac { \sqrt { 2 } }{ 2 } \tanh ^{ -1 }{ \frac { (u+\frac { 1 }{ u } )\sqrt { 2 } }{ 2 } } +{ c } \\ \quad & = \boxed { \frac { \sqrt { 2 } }{ 2 } (\tan ^{ -1 }{ \frac { (\sqrt { \tan { x } } -\sqrt { \cot { x } } )\sqrt { 2 } }{ 2 } } -\tanh ^{ -1 }{ \frac { (\sqrt { \tan { x } } +\sqrt { \cot { x } } )\sqrt { 2 } }{ 2 } } )+c } \end{aligned}

Note by Gordon Chan
4 months, 3 weeks ago

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Wait, is tanh really involved?

Ruilin Wang - 2 months, 3 weeks ago

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Yes.

Gordon Chan - 2 months, 3 weeks ago

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