14. Integration of the square-root of tan x

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Consider $$ \begin{align} I = \int \sqrt{\tan x}\,\textrm{d}x \quad \textrm{for} \quad 0 \le x < \frac{\pi}{2} \end{align} $$

 

(a) Method 1. By using the substitution $u = \sqrt{ \tan x }$, followed by partial fractions, we can show that $$ \begin{align} I &= \frac1{2\sqrt{2}} \ln \left\vert \frac{ \tan x - \sqrt{2 \tan x} + 1 }{ \tan x + \sqrt{2 \tan x} + 1 } \right\vert + \frac1{\sqrt{2}}\arctan \left( \frac{ \tan x - 1 }{\sqrt{2 \tan x}} \right) + C \\ \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{ \sqrt{\tan x} + \sqrt{\cot x} - \sqrt{2} }{ \sqrt{\tan x} + \sqrt{\cot x} + \sqrt{2} } \right\vert + \frac1{\sqrt{2}}\arctan \left( \frac{\sqrt{\tan x} - \sqrt{\cot x}}{\sqrt{2}} \right) + C \end{align} $$

 

(b) Method 2. By using the substitution $u = \sqrt{ \tan x }$, followed by two separate substitutions $z = u + \frac1u$ and $t = u - \frac1u$, we can derive the same result as (a).

 

(c) Method 3. By considering $I$ together with $J=\sqrt{\cot x}$, we can show that $$ \begin{align} I &= \frac1{\sqrt{2}}\arcsin \left( \sin x - \cos x \right) - \frac1{\sqrt{2}} \textrm{arcosh} (\sin x + \cos x) + C \\ &= \frac1{\sqrt{2}}\arcsin \left[ \sqrt{2} \sin \left( x - \frac{\pi}{4} \right) \right] - \frac1{\sqrt{2}} \textrm{arcosh} \left[ \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \right] + C \\ &= \frac1{\sqrt{2}}\arcsin \left( \sin x - \cos x \right) - \frac1{\sqrt{2}} \ln \left( \sin x + \cos x + \sqrt{ \sin 2x } \right) + C \end{align} $$

 

Solution.

 

(a) Method 1. Substitution $u = \sqrt{ \tan x }$, followed by partial fractions.

 

Step 1. Use the substitution, $$ \begin{align} u &= \sqrt{ \tan x } = ( \tan x )^{\frac12} \quad \left( u \ge 0 \right) \\ \Rightarrow\qquad \frac{ \textrm{d}u }{ \textrm{d}x } &= \frac12( \tan x )^{-\frac12} \sec^2x = \frac{1 + \tan^2x}{2\sqrt{ \tan x}} = \frac{1 + u^4}{2u} \end{align} $$ and the integral becomes: $$ \begin{align} I &= \int \left( \sqrt{\tan x} \, \frac{ \textrm{d}x }{ \textrm{d}u } \right) \textrm{d}u \\ &= \int \left( u \times \frac{2u}{1 + u^4} \right) \textrm{d}u \\ &= \int \frac{2u^2}{1 + u^4}\,\textrm{d}u \end{align} $$

 

Step 2. We try to factorise the denominator. For $1 + u^4$, we can think of the following two possibilities. $$ \begin{align} \textrm{(1)}&& \left(1 + au + u^2 \right)\left(1 + bu + u^2 \right) &= 1 + (a+b)u + (ab+2)u^2 + (a+b)u^3 + u^4 \\ \textrm{(2)}&& \left(1 + au - u^2 \right)\left(1 + bu - u^2 \right) &= 1 + (a+b)u + (ab-2)u^2 - (a+b)u^3 + u^4 \end{align} $$ The linear and cubic terms require $a+b=0$. The quadratic term then reads $\left( - a^2 \pm 2 \right) u^2$ and, for $a\in\mathbb R$, only the positive sign is possible, i.e. we choose (1) with $a=-b=\sqrt{2}$, i.e. $$ \begin{align} 1 + u^4 = \left(1 - \sqrt{2}u + u^2 \right)\left(1 + \sqrt{2}u + u^2 \right) \end{align} $$ The integrand then reads by partial fractions: $$ \begin{align} \frac{2u}{\left(1 - \sqrt{2}u + u^2 \right)\left(1 + \sqrt{2}u + u^2 \right)} &= \frac{A+Bu}{1 - \sqrt{2}u + u^2} + \frac{C+Du}{1 + \sqrt{2}u + u^2} \\ &= \frac{ (A+Bu) \left(1 + \sqrt{2}u + u^2 \right) + (C+Du) \left(1 - \sqrt{2}u + u^2 \right) }{\left(1 - \sqrt{2}u + u^2 \right)\left(1 + \sqrt{2}u + u^2 \right)} \\ &= \frac{ (A+C) + [\sqrt{2}(A-C) + (B+D)]u + [(A+C) + \sqrt{2}(B-D)]u^2 + (B+D)u^3 }{\left(1 - \sqrt{2}u + u^2 \right)\left(1 + \sqrt{2}u + u^2 \right)} \\ \end{align} $$ The constant and cubic terms require $A+C=0$ and $B+D=0$. The linear term then sets $A=C=0$ and the quadratic term gives $B=-D=\frac1{\sqrt{2}}$.

 

Step 3. $$ \begin{align} \Rightarrow\qquad I &= \int \frac{2u}{\left(1 - \sqrt{2}u + u^2 \right)\left(1 + \sqrt{2}u + u^2 \right)}\, \textrm{d}u \\ &= \int \left( \frac{\frac1{\sqrt{2}}u}{1 - \sqrt{2}u + u^2} - \frac{\frac1{\sqrt{2}}u}{1 + \sqrt{2}u + u^2} \right) \textrm{d}u \quad (\textrm{by partial fractions})\\ &= \int \frac1{2\sqrt{2}}\left( \frac{2u}{1 - \sqrt{2}u + u^2} - \frac{2u}{1 + \sqrt{2}u + u^2} \right) \textrm{d}u \\ &= \int \frac1{2\sqrt{2}}\left( \frac{\left( -\sqrt{2} + 2u \right) + \sqrt{2}}{1 - \sqrt{2}u + u^2} - \frac{\left( \sqrt{2} + 2u \right) - \sqrt{2}}{1 + \sqrt{2}u + u^2} \right) \textrm{d}u \\ &= \int \frac1{2\sqrt{2}}\left( \frac{ -\sqrt{2} + 2u }{1 - \sqrt{2}u + u^2} + \frac{\sqrt{2}}{1 - \sqrt{2}u + u^2} - \frac{ \sqrt{2} + 2u }{1 + \sqrt{2}u + u^2} + \frac{\sqrt{2}}{1 + \sqrt{2}u + u^2} \right) \textrm{d}u \\ &= \frac1{2\sqrt{2}} \Big( \ln \left\vert 1 - \sqrt{2}u + u^2 \right\vert - \ln \left\vert 1 + \sqrt{2}u + u^2 \right\vert \Big) + \int \frac12 \left( \frac{1}{1 - \sqrt{2}u + u^2} + \frac{1}{1 + \sqrt{2}u + u^2} \right) \textrm{d}u \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{1 - \sqrt{2}u + u^2}{1 + \sqrt{2}u + u^2} \right\vert + \frac12 \int \left( \frac{1}{\left( u - \frac{1}{\sqrt{2}} \right)^2+\frac12 } + \frac{1}{ \left( u + \frac{1}{\sqrt{2}} \right)^2+\frac12 } \right) \textrm{d}u \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{1 - \sqrt{2}u + u^2}{1 + \sqrt{2}u + u^2} \right\vert + \frac12 \cdot \sqrt{2}\arctan \left[ \sqrt{2} \left( u - \frac1{\sqrt{2}} \right) \right] + \frac12 \cdot \sqrt{2}\arctan \left[ \sqrt{2} \left( u + \frac1{\sqrt{2}} \right) \right] + C' \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{1 - \sqrt{2}u + u^2}{1 + \sqrt{2}u + u^2} \right\vert + \frac1{\sqrt{2}} \left[ \arctan \left( \sqrt{2}u - 1 \right) + \arctan \left( \sqrt{2}u + 1 \right) \right] + C' \end{align} $$ since $$ \begin{align} \int \frac1{(x-a)^2+b^2} \, \textrm{d}x = \frac1{b}\arctan\left( \frac{x-a}{b} \right) + C \end{align} $$

 

Step 4. We can simplify the final expression slightly further by recalling: $$ \begin{align} &&& \arctan x + \arctan y = \left\{ \begin{array}{lllll} \arctan \left( \frac{ x + y }{ 1 - xy } \right) + \pi, & xy > 1, & x>0, & y>0 & \left( y > \frac1x \right) \\ \arctan \left( \frac{ x + y }{ 1 - xy } \right) - \pi, & xy > 1, & x<0, & y<0 & \left( y < \frac1x \right) \\ \arctan \left( \frac{ x + y }{ 1 - xy } \right), & \textrm{otherwise} \end{array} \right. \\ \\ &\Rightarrow& & \arctan \left( \sqrt{2} u - 1 \right) + \arctan \left( \sqrt{2} u + 1 \right) = \arctan \left[ \frac{ \left( \sqrt{2} u - 1 \right) + \left( \sqrt{2} u + 1 \right) }{ 1 - \left( \sqrt{2} u - 1 \right) \left( \sqrt{2} u + 1 \right) } \right] = \arctan \left( \frac{ \sqrt{2} u }{ 1 - u^2 } \right) \\ \\ &\Rightarrow& & \arctan \left( \sqrt{2} u - 1 \right) + \arctan \left( \sqrt{2} u + 1 \right) = \left\{ \begin{array}{lcl} \arctan \left( \frac{ \sqrt{2} u }{ 1 - u^2 } \right) + \pi & \textrm{for} & u > 1 \\ \arctan \left( \frac{ \sqrt{2} u }{ 1 - u^2 } \right) - \pi & \textrm{for} & u < - 1 \\ \arctan \left( \frac{ \sqrt{2} u }{ 1 - u^2 } \right) & \textrm{for} & -1 \le u \le 1 \end{array} \right. \end{align} $$

 

which gives $$ \begin{align} \arctan \left( \sqrt{2}u - 1 \right) + \arctan \left( \sqrt{2}u + 1 \right) &= \arctan \left( \frac{ u^2 - 1 }{ \sqrt{2}u } \right) + \textrm{sgn}(u)\frac{\pi}{2} \end{align} $$ where, for the last step, we used $$ \begin{align} \arctan x + \textrm{arccot}\, x &= \frac{\pi}{2} \\ \Leftrightarrow\qquad \arctan x + \arctan \left(\frac1x\right) &= \left\{ \begin{array}{ll} \frac{\pi}{2}, & x > 0 \\ -\frac{\pi}{2}, & x < 0 \end{array} \right. \end{align} $$ and $\tan(-x)=-\tan x$.

 

Finally, noting that $u = \sqrt{\tan x} \ge 0$, we find: $$ \begin{align} \Rightarrow\qquad I &= \frac1{2\sqrt{2}} \ln \left\vert \frac{1 - \sqrt{2}u + u^2}{1 + \sqrt{2}u + u^2} \right\vert + \frac1{\sqrt{2}} \arctan \left( \frac{ u^2 - 1 }{ \sqrt{2}u } \right) + \underbrace{\textrm{sgn}(u)\frac{\pi}{2\sqrt{2}} + C'}_{=C} \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{1 - \sqrt{2 \tan x} + \tan x}{1 + \sqrt{2 \tan x} + \tan x} \right\vert + \frac1{\sqrt{2}} \arctan \left( \frac{ \tan x - 1 }{ \sqrt{2 \tan x} } \right) + C \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{ \sqrt{\tan x} + \sqrt{\cot x} - \sqrt{2} }{ \sqrt{\tan x} + \sqrt{\cot x} + \sqrt{2} } \right\vert + \frac1{\sqrt{2}} \arctan \left( \frac{ \sqrt{\tan x} - \sqrt{\cot x} }{ \sqrt{2} } \right) + C \qquad \square \end{align} $$

 

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(b) Method 2. Substitution $u = \sqrt{ \tan x }$, followed by two separate substitutions.

 

Step 1. Use the substitution, $$ \begin{align} u &= \sqrt{ \tan x } = ( \tan x )^{\frac12} \quad \left( 0 \le x < \frac{\pi}{2}, u \ge 0 \right) \\ \Rightarrow\qquad \frac{ \textrm{d}u }{ \textrm{d}x } &= \frac12( \tan x )^{-\frac12} \sec^2x = \frac{1 + \tan^2x}{2\sqrt{ \tan x}} = \frac{1 + u^4}{2u} \end{align} $$ and the integral becomes: $$ \begin{align} I &= \int \left( \sqrt{\tan x} \, \frac{ \textrm{d}x }{ \textrm{d}u } \right) \textrm{d}u \\ &= \int \left( u \times \frac{2u}{1 + u^4} \right) \textrm{d}u \\ &= \int \frac{2u^2}{1 + u^4} \,\textrm{d}u \\ &= \int \frac{2}{u^2 + \frac1{u^2}} \,\textrm{d}u \\ &= \int \left[ \frac{1 - \frac1{u^2} + 1 + \frac1{u^2}}{u^2 + \frac1{u^2}} \right] \textrm{d}u \\ &= \int \left[ \frac{1 - \frac1{u^2}}{ \left( u + \frac1{u} \right)^2 - 2} + \frac{1 + \frac1{u^2}}{ \left( u - \frac1{u} \right)^2 + 2} \right] \textrm{d}u \\ \end{align} $$

 

Step 2. Use the substitutions (for $u\ge 0$): $$ \begin{align} z &= u + \frac1u \quad (z \ge 2) &&\textrm{and}& t &= u - \frac1u \quad (t\in\mathbb R) \\ \textrm{d}z &= \left( 1 - \frac1{u^2} \right) \textrm{d}u &&\textrm{and}& \textrm{d}t &= \left( 1 + \frac1{u^2} \right) \textrm{d}u \end{align} $$ The integral becomes $$ \begin{align} \Rightarrow\qquad I &= \int \frac{ 1 }{ z^2 - 2 } \, \textrm{d}z + \int \frac{ 1 }{ t^2 + 2 } \, \textrm{d}t \\ &= \int \frac{ 1 }{ \left( z - \sqrt{2} \right) \left( z + \sqrt{2} \right) } \, \textrm{d}z + \frac1{\sqrt{2}}\arctan \left( \frac{t}{\sqrt{2}} \right) + C \\ &= \int \frac1{2\sqrt{2}} \left( \frac{ 1 }{ z - \sqrt{2} } - \frac{ 1 }{ z + \sqrt{2} } \right) \, \textrm{d}z + \frac1{\sqrt{2}}\arctan \left( \frac{t}{\sqrt{2}} \right) + C \\ &= \frac1{2\sqrt{2}} \Big( \ln \left\vert z - \sqrt{2} \right\vert - \ln \left\vert z + \sqrt{2} \right\vert \Big) + \frac1{\sqrt{2}}\arctan \left( \frac{t}{\sqrt{2}} \right) + C' \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{ z - \sqrt{2} }{ z + \sqrt{2} } \right\vert + \frac1{\sqrt{2}}\arctan \left( \frac{t}{\sqrt{2}} \right) + C' \end{align} $$

 

since $$ \begin{align} \int \frac1{x^2 - a^2} \, \textrm{d}x &= - \frac1{a} \textrm{arcoth}\left( \frac{x}{a} \right) + C = \frac1{2a} \ln \left( \frac{x-a}{x+a} \right) + C \quad (|x|>a) \\ \int \frac1{a^2 - x^2} \, \textrm{d}x &= \frac1{a} \textrm{artanh}\left( \frac{x}{a} \right) + C = \frac1{2a} \ln \left( \frac{a+x}{a-x} \right) + C \quad (-a < x < a) \\ \int \frac1{x^2 + a^2} \, \textrm{d}x &= \frac1{a} \arctan \left( \frac{x}{a} \right) + C \quad (x \in \mathbb R) \end{align} $$ Finally, $$ \begin{align} I &= \frac1{2\sqrt{2}} \ln \left\vert \frac{ u + \frac1u - \sqrt{2} }{ u + \frac1u + \sqrt{2} } \right\vert + \frac1{\sqrt{2}}\arctan \left( \frac{u - \frac1u}{\sqrt{2}} \right) + C \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{ \sqrt{\tan x} + \sqrt{\cot x} - \sqrt{2} }{ \sqrt{\tan x} + \sqrt{\cot x} + \sqrt{2} } \right\vert + \frac1{\sqrt{2}}\arctan \left( \frac{\sqrt{\tan x} - \sqrt{\cot x}}{\sqrt{2}} \right) + C \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{ \tan x - \sqrt{2 \tan x} + 1 }{ \tan x + \sqrt{2 \tan x} + 1 } \right\vert + \frac1{\sqrt{2}}\arctan \left( \frac{ \tan x - 1 }{\sqrt{2 \tan x}} \right) + C \qquad \square \end{align} $$

 

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(c) Method 3. We consider the following two integrals: $$ \begin{align} I &= \int \sqrt{ \tan x }\, \textrm{d}x,\quad \left( 0 \le x < \frac{\pi}{2} \right) \\ J &= \int \sqrt{ \cot x }\, \textrm{d}x,\quad \left( 0 < x \le \frac{\pi}{2} \right) \end{align} $$ Now, consider $I+J$ and $I-J$.

 

Step 1. For $I+J$, $$ \begin{align} I+J &= \int \left( \sqrt{ \tan x } + \sqrt{ \cot x } \right) \textrm{d}x \\ &= \int \left( \frac{ \sqrt{\sin x} }{ \sqrt{\cos x} } + \frac{ \sqrt{\cos x} }{ \sqrt{\sin x} } \right) \textrm{d}x \\ &= \int \frac{ \sin x + \cos x }{ \sqrt{ \sin x \cos x } } \, \textrm{d}x \\ &= \sqrt{2} \int \frac{ \sin x + \cos x }{ \sqrt{ 2 \sin x \cos x } } \, \textrm{d}x \end{align} $$ We note that: $$ \begin{align} ( \sin x + \cos x )^2 &= 1 + 2 \sin x \cos x \\ ( \sin x - \cos x )^2 &= 1 - 2 \sin x \cos x \\ \\ \Rightarrow\qquad 2 \sin x \cos x &= ( \sin x + \cos x )^2 - 1 \\ &= 1 - ( \sin x - \cos x )^2 \end{align} $$ and that $\textrm{d}( \sin x - \cos x ) = ( \cos x + \sin x )\textrm{d}x$, $I+J$ can be written as: $$ \begin{align} \Rightarrow\qquad I+J &= \sqrt{2} \int \frac{ \sin x + \cos x }{ \sqrt{ 1 - ( \sin x - \cos x )^2 } } \, \textrm{d}x \end{align} $$ Use the substitution: $$ \begin{align} u &= \sin x - \cos x \\ \textrm{d}u &= \left( \cos x + \sin x \right) \textrm{d}x \end{align} $$ and the integral becomes: $$ \begin{align} \Rightarrow\qquad I+J &= \sqrt{2} \int \frac{ 1 }{ \sqrt{ 1 - u^2 } } \, \textrm{d}u \\ &= \sqrt{2} \arcsin u + C' \\ &= \sqrt{2} \arcsin \left( \sin x - \cos x \right) + C' \end{align} $$

 

Step 2. For $I-J$, $$ \begin{align} I-J &= \int \left( \sqrt{ \tan x } - \sqrt{ \cot x } \right) \textrm{d}x \\ &= \int \left( \frac{ \sqrt{\sin x} }{ \sqrt{\cos x} } - \frac{ \sqrt{\cos x} }{ \sqrt{\sin x} } \right) \textrm{d}x \\ &= \int \frac{ \sin x - \cos x }{ \sqrt{ \sin x \cos x } } \, \textrm{d}x \\ &= \sqrt{2} \int \frac{ \sin x - \cos x }{ \sqrt{ 2 \sin x \cos x } } \, \textrm{d}x \end{align} $$ Recall: $$ \begin{align} 2 \sin x \cos x &= ( \sin x + \cos x )^2 - 1 \\ &= 1 - ( \sin x - \cos x )^2 \end{align} $$ and note that $\textrm{d}( \sin x + \cos x ) = ( \cos x - \sin x )\textrm{d}x$, $I-J$ can be written as: $$ \begin{align} \Rightarrow\qquad I-J &= \sqrt{2} \int \frac{ \sin x + \cos x }{ \sqrt{ ( \sin x + \cos x )^2 - 1 } } \, \textrm{d}x \end{align} $$ Use the substitution: $$ \begin{align} u &= \sin x + \cos x \\ \textrm{d}u &= \left( \cos x - \sin x \right) \textrm{d}x \end{align} $$ and the integral becomes: $$ \begin{align} \Rightarrow\qquad I-J &= - \sqrt{2} \int \frac{ 1 }{ \sqrt{ u^2 - 1 } } \, \textrm{d}u \\ &= - \sqrt{2} \, \textrm{arcosh}\, u + C'' \\ &= - \sqrt{2} \, \textrm{arcosh}(\sin x + \cos x) + C'' \\ &= - \sqrt{2} \ln \left( u + \sqrt{ u^2 - 1 } \right) + C'' \\ &= - \sqrt{2} \ln \left( \sin x + \cos x + \sqrt{ (\sin x + \cos x)^2 - 1 } \right) + C'' \\ &= - \sqrt{2} \ln \left( \sin x + \cos x + \sqrt{ 2 \sin x \cos x } \right) + C'' \\ &= - \sqrt{2} \ln \left( \sin x + \cos x + \sqrt{ \sin 2x } \right) + C'' \end{align} $$

 

Step 3. We collect the results: $$ \begin{align} I+J &= \sqrt{2} \arcsin \left( \sin x - \cos x \right) + C' \\ I-J &= - \sqrt{2} \, \textrm{arcosh}(\sin x + \cos x) + C'' = - \sqrt{2} \ln \left( \sin x + \cos x + \sqrt{ \sin 2x } \right) + C'' \end{align} $$ and they yield: $$ \begin{align} I = \int \sqrt{\tan x}\,\textrm{d}x &= \frac1{\sqrt{2}}\arcsin \left( \sin x - \cos x \right) - \frac1{\sqrt{2}} \textrm{arcosh} (\sin x + \cos x) + C \\ &= \frac1{\sqrt{2}}\arcsin \left[ \sqrt{2} \sin \left( x - \frac{\pi}{4} \right) \right] - \frac1{\sqrt{2}} \textrm{arcosh} \left[ \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \right] + C \\ &= \frac1{\sqrt{2}}\arcsin \left( \sin x - \cos x \right) - \frac1{\sqrt{2}} \ln \left( \sin x + \cos x + \sqrt{ \sin 2x } \right) + C \\ \\ J = \int \sqrt{\cot x}\,\textrm{d}x &= \frac1{\sqrt{2}}\arcsin \left( \sin x - \cos x \right) + \frac1{\sqrt{2}} \, \textrm{arcosh}(\sin x + \cos x) + C \\ &= \frac1{\sqrt{2}}\arcsin \left[ \sqrt{2} \sin \left( x - \frac{\pi}{4} \right) \right] + \frac1{\sqrt{2}} \, \textrm{arcosh}\left[ \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \right] + C \\ &= \frac1{\sqrt{2}}\arcsin \left( \sin x - \cos x \right) + \frac1{\sqrt{2}} \ln \left( \sin x + \cos x + \sqrt{ \sin 2x } \right) + C \end{align} $$

 

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Aside. Together with the results for $J=\int \sqrt{\cot x} \, \textrm{d}x$ (see here), we can make a comparison:

 

$$ \begin{align} I = \int \sqrt{\tan x}\,\textrm{d}x &= \frac1{\sqrt{2}}\arcsin \left( \sin x - \cos x \right) - \frac1{\sqrt{2}} \textrm{arcosh} (\sin x + \cos x) + C \\ \\ &= \frac1{\sqrt{2}}\arcsin \left[ \sqrt{2} \sin \left( x - \frac{\pi}{4} \right) \right] - \frac1{\sqrt{2}} \textrm{arcosh} \left[ \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \right] + C \\ \\ &= \frac1{\sqrt{2}}\arcsin \left( \sin x - \cos x \right) - \frac1{\sqrt{2}} \ln \left( \sin x + \cos x + \sqrt{ \sin 2x } \right) + C \\ \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{ \tan x - \sqrt{2 \tan x} + 1 }{ \tan x + \sqrt{2 \tan x} + 1 } \right\vert + \frac1{\sqrt{2}}\arctan \left( \frac{ \tan x - 1 }{\sqrt{2 \tan x}} \right) + C \\ \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{ \sqrt{\tan x} + \sqrt{\cot x} - \sqrt{2} }{ \sqrt{\tan x} + \sqrt{\cot x} + \sqrt{2} } \right\vert + \frac1{\sqrt{2}}\arctan \left( \frac{\sqrt{\tan x} - \sqrt{\cot x}}{\sqrt{2}} \right) + C \end{align} $$

 

$$ \begin{align} J = \int \sqrt{\cot x}\,\textrm{d}x &= \frac1{\sqrt{2}}\arcsin \left( \sin x - \cos x \right) + \frac1{\sqrt{2}} \, \textrm{arcosh}(\sin x + \cos x) + C \\ \\ &= \frac1{\sqrt{2}}\arcsin \left[ \sqrt{2} \sin \left( x - \frac{\pi}{4} \right) \right] + \frac1{\sqrt{2}} \, \textrm{arcosh}\left[ \sqrt{2} \sin \left( x + \frac{\pi}{4} \right) \right] + C \\ \\ &= \frac1{\sqrt{2}}\arcsin \left( \sin x - \cos x \right) + \frac1{\sqrt{2}} \ln \left( \sin x + \cos x + \sqrt{ \sin 2x } \right) + C \\ \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{ \cot x - \sqrt{2 \cot x} + 1 }{ \cot x + \sqrt{2 \cot x} + 1 } \right\vert + \frac1{\sqrt{2}}\textrm{arccot} \left( \frac{ \cot x - 1 }{\sqrt{2 \cot x}} \right) + C \\ \\ &= \frac1{2\sqrt{2}} \ln \left\vert \frac{ \sqrt{\cot x} + \sqrt{\tan x} + \sqrt{2} }{ \sqrt{\cot x} + \sqrt{\tan x} - \sqrt{2} } \right\vert + \frac1{\sqrt{2}}\textrm{arccot} \left( \frac{\sqrt{\cot x} - \sqrt{\tan x}}{\sqrt{2}} \right) + C \end{align} $$