\(\int { \sqrt { \tan { x } } } \, dx \)

∫tan⁡x dx=u=lettan⁡x∫2u2u4+1du=∫2u2+1u2du=∫1+1u2(u−1u)2+2du+∫1−1u2(u+1u)2−2du=v=letu−1uw=letu+1u∫1v2+2dv+∫1w2−2dw=12tan⁡−1v2+c1−12tanh⁡−1w2+c2=22tan⁡−1(u−1u)22−22tanh⁡−1(u+1u)22+c=22(tan⁡−1(tan⁡x−cot⁡x)22−tanh⁡−1(tan⁡x+cot⁡x)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}

#Calculus

Note by Gandoff Tan
2 years ago

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Comments

I finally found the answer to the hardest integral

Jay Mark Acedo - 1 year, 7 months ago

Wait, is tanh really involved?

Ruilin Wang - 1 year, 10 months ago

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Yes.

Gandoff Tan - 1 year, 10 months ago
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