# $\int { \sqrt { \tan { x } } } \, dx$

\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 Gandoff Tan
2 years, 2 months ago

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I finally found the answer to the hardest integral

- 1 year, 9 months ago

Wait, is tanh really involved?

- 2 years ago

Yes.

- 2 years ago