Question e882e

Explanation:

Credits:
1. Thanks to the web site www.integral-calculator.com that gave us the direction to be followed to solve this integral!
2. Thanks to the website www.symbolab.com that reminded us how to express complex roots in the form a + bi!

Mar 2, 2018

$\int \setminus \setminus \sqrt{\tan \left(x\right)} \setminus \mathrm{dx} \setminus =$

 = \sqrt{ 2 }/4 ( ln| { tan x - \sqrt{ 2 tan x } + 1 }/{ tan x + \sqrt{ 2 tan x } + 1 } | + 2 arctan( { \sqrt{ 2 tan x } }/{ 1 - tan x } ) ) + C. 

$\text{Please see solution below. Maybe simple ??!}$

Explanation:

$\text{We want to look at the integral:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \int \setminus \sqrt{\tan \left(x\right)} \setminus \mathrm{dx} .$

$\text{When someone asked me this many years ago, a substitution}$
$\text{occurred to me then, and resolved the situation, without too}$
$\text{much difficulty. The calculus was fairly direct; there was some}$
$\text{post-calculus algebra that supeficially appeared complex, but}$
$\text{structurally was not. Later, I found that slightly adjusting}$
$\text{this substitution eliminates some of the superficial complexity}$
$\text{of the algebra involved with the simplification.}$

 "The substitution I originally used was:" \qquad u \ = \ sqrt{ tan(x) }; 

$\text{The one I found to be a bit better is:} \setminus q \quad \setminus q \quad u \setminus = \setminus \sqrt{2 \tan \left(x\right)} .$

$\text{So, let's go ahead with this substitution:} \setminus q \quad \setminus u \setminus = \setminus \sqrt{2 \tan \left(x\right)} .$

$\text{1) Let:} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad u \setminus = \setminus \sqrt{2 \tan \left(x\right)} . \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad \setminus \setminus \setminus \left(I\right)$

$\text{2) So:} \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \mathrm{du} = \frac{{\sec}^{2} \left(x\right)}{\sqrt{2 \tan \left(x\right)}} \setminus \mathrm{dx} .$

$\text{3) Thus:} \setminus q \quad \setminus q \quad \setminus q \quad \mathrm{dx} = \frac{\sqrt{2 \tan \left(x\right)}}{{\sec}^{2} \left(x\right)} \setminus \mathrm{du} .$

$\text{4) So, we begin (!!):}$

$\setminus q \quad \setminus q \quad \setminus \quad \int \setminus \sqrt{\tan \left(x\right)} \setminus \mathrm{dx}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus \int \setminus \sqrt{\tan \left(x\right)} \cdot \frac{\sqrt{2 \tan \left(x\right)}}{{\sec}^{2} \left(x\right)} \setminus \mathrm{du}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus \sqrt{2} \int \setminus \tan \frac{x}{\sec} ^ 2 \left(x\right) \setminus \mathrm{du}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus \sqrt{2} \setminus \int \setminus \tan \frac{x}{1 + {\tan}^{2} \left(x\right)} \setminus \mathrm{du}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus \sqrt{2} \setminus \int \setminus \frac{\frac{1}{2} {u}^{2}}{1 + \frac{1}{4} {u}^{4}} \setminus \mathrm{du} \setminus q \quad \setminus q \quad \setminus q \quad \left[\text{as} \setminus \setminus \tan \left(x\right) \setminus = \setminus \frac{1}{2} {u}^{2}\right]$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus \sqrt{2} \setminus \int \setminus \frac{4}{4} \cdot \frac{\frac{1}{2} {u}^{2}}{1 + \frac{1}{4} {u}^{4}} \setminus \mathrm{du}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus 2 \sqrt{2} \setminus \int \setminus {u}^{2} / \left\{4 + {u}^{4}\right\} \setminus \mathrm{du} .$

$\setminus q \quad \setminus q \quad \therefore \setminus q \quad \setminus q \quad \int \setminus \sqrt{\tan \left(x\right)} \setminus \mathrm{dx} \setminus = \setminus 2 \sqrt{2} \setminus \int \setminus {u}^{2} / \left\{4 + {u}^{4}\right\} \setminus \mathrm{du} . \setminus q \quad \setminus q \quad \setminus q \quad \setminus \left(I I\right)$

$\text{At this point, the essential calculus is over. The remaining}$
$\text{calculus is the method of partial fractions -- deterministic.}$
$\text{Any complexity from this point on -- superficial or structural,}$
$\text{is purely algebraic, and not analytic.}$

$\text{5) Expand integrand into partial fractions:}$

$\setminus q \quad \text{a) Factor denominator:}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad {u}^{4} + 4 \setminus = \setminus {u}^{4} + 4 {u}^{2} + 4 - 4 {u}^{2}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus {\left({u}^{2} + 2\right)}^{2} - {\left(2 u\right)}^{2}$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus \left[{u}^{2} + 2 - 2 u\right] \cdot \left[{u}^{2} + 2 + 2 u\right]$

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus \left[{u}^{2} - 2 u + 2\right] \cdot \left[{u}^{2} + 2 u + 2\right] .$

$\setminus q \quad \setminus q \quad \text{Can tell the two factors on the RHS of the previous are}$
$\text{irreducible over" \ \ RR, "because their discriminants are both}$
$\text{negative. So:}$

$\setminus q \quad \setminus q \quad \text{Factorization of denominator, over} \setminus \mathbb{R} :$

 \qquad \qquad \qquad \qquad \quad \ u^4 + 4 \ = \ ( u^2 - 2 u + 2 ) ( u^2 + 2 u + 2 ). 

$\setminus q \quad \text{b) Partial Fraction Decomposition of integrand:}$

 \qquad \qquad \qquad \qquad \ u^2/{ 4 + u^4 } \ = \ { a u + b }/{ u^2 - 2 u + 2 } + { c u + d }/{ u^2 + 2 u + 2 }. 

$\setminus q \quad \setminus q \quad \text{Using the standard technique of clearing the}$
$\text{polynomial fractions, and equating coefficients of like powers}$
$\text{of" \ \ u \ \ "in the resulting equality of polynomials, and solving}$
$\text{the resulting linear system for" \ \ a, b, c, d \ \ "we find:}$

$\setminus q \quad \setminus q \quad \setminus q \quad a = \frac{1}{4} \setminus q \quad \setminus q \quad b = 0 \setminus q \quad \setminus q \quad c = - \frac{1}{4} \setminus q \quad \setminus q \quad d = 0.$

$\text{[These values can easily be checked in the decomposition}$
$\text{equality above -- almost by eye alone !]}$

$\setminus q \quad \setminus q \quad \text{So, the partial fraction decomposition of the integrand is:}$

 \qquad \qquad \qquad u^2/{ 4 + u^4 } \ = \ 1/4 ( u/{ u^2 - 2 u + 2 } - u/( u^2 + 2 u + 2 ) ). 

$\text{6) Completion of Integration:}$

$\setminus q \quad \setminus q \quad \text{Now we complete the integration, following from the}$
$\text{partial fraction decomposition, using the standard technique of}$
$\text{adjusting the polynomial fractions there, to set up a ln part,}$
$\text{and an arctan part:}$

 \qquad \qquad \qquad u^2/{ 4 + u^4 } \ = \ 1/4 ( u/{ u^2 - 2 u + 2 } - u/{ u^2 + 2 u + 2 } )

 \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 1/4 cdot 1/2 ( { 2 u }/{ u^2 - 2 u + 2 } - { 2 u }/{ u^2 + 2 u + 2 } )

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \quad = \setminus \frac{1}{8} \left(\frac{2 u - 2 + 2}{{u}^{2} - 2 u + 2} - \frac{2 u + 2 - 2}{{u}^{2} + 2 u + 2}\right)$

 \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 1/8 ( { 2 u - 2 }/{ u^2 - 2 u + 2 } + { 2 }/{ u^2 - 2 u + 2 } 

 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ - { 2 u + 2 }/{ u^2 + 2 u + 2 } + { 2 }/{ u^2 + 2 u + 2 } ) 

\qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 1/8 ( { 2 u - 2 }/{ u^2 - 2 u + 2 } + { 2 }/{ ( u - 1 )^2 + 1 } 

 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad - { 2 u + 2 }/{ u^2 + 2 u + 2 } + { 2 }/{ ( u + 1 )^2 + 1 } ) 

 \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 1/8 ( \underbrace{ { 2 u - 2 }/{ u^2 - 2 u + 2 } - { 2 u + 2 }/{ u^2 + 2 u + 2 }}_{ "ln part"} 

 \qquad \qquad \qquad \qquad \qquad \qquad \qquad + \underbrace{ { 2 }/{ ( u - 1 )^2 + 1 }+ { 2 }/{ ( u + 1 )^2 + 1 } }_{ "arctan part"} ). \qquad \qquad (III) 

$\setminus q \quad \setminus q \quad \text{Now by (II), and (III), we have, altogether:}$

$\setminus \int \setminus \sqrt{\tan \left(x\right)} \setminus \mathrm{dx} \setminus = \setminus 2 \setminus \sqrt{2} \setminus \int \setminus {u}^{2} / \left\{4 + {u}^{4}\right\} \setminus \mathrm{du}$

 \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ 2 \sqrt{ 2 } int \ 1/8 ( { 2 u - 2 }/{ u^2 - 2 u + 2 } - { 2 u + 2 }/{ u^2 + 2 u + 2 } 

 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ + { 2 }/{ ( u - 1 )^2 + 1 }+ { 2 }/{ ( u + 1 )^2 + 1 } ) \ du 

 \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ { 2 \sqrt{ 2 } }/8 int \ ( { 2 u - 2 }/{ u^2 - 2 u + 2 } - { 2 u + 2 }/{ u^2 + 2 u + 2 } 

 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ \ + { 2 }/{ ( u - 1 )^2 + 1 }+ { 2 }/{ ( u + 1 )^2 + 1 } ) \ du 

 \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ \sqrt{ 2 }/4 ( ln| u^2 - 2 u + 2 | - ln | u^2 + 2 u + 2 | 

 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ +2 arctan( u - 1 )+ 2 arctan( u + 1 ) )+ C 

 = \ \sqrt{ 2 }/4 ( ln| { u^2 - 2 u + 2 }/{ u^2 + 2 u + 2 } | + 2 arctan[ { ( u - 1 ) + ( u + 1 ) }/{ 1 - ( u - 1 )( u + 1 ) } ] ) + C 

 = \ \sqrt{ 2 }/4 ( ln| { u^2 - 2 u + 2 }/{ u^2 + 2 u + 2 } | + 2 arctan( { 2 u }/{ 2 - u^2 } ) )+ C 

$\setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \left[\setminus \text{as, by (I):} \setminus \quad u = \setminus \sqrt{2 \tan \left(x\right)} \setminus q \quad \Rightarrow \setminus q \quad\right]$

 = \sqrt{ 2 }/4 [ ln| { 2 tan x - 2 \sqrt{ 2 tan x } + 2 }/{ 2 tan x + 2 \sqrt{ 2 tan x } + 2} | + 2 arctan( { 2 \sqrt{ 2 tan x } }/{ 2 - 2 tan x } ) ] + C 

 = \sqrt{ 2 }/4 [ ln| { tan x - \sqrt{ 2 tan x } + 1 }/{ tan x + \sqrt{ 2 tan x } + 1 } | + 2 arctan( { \sqrt{ 2 tan x } }/{ 1 - tan x } ) ] + C. 

$\setminus q \quad \setminus q \quad \text{So, we now have here:}$

$\setminus \int \setminus \sqrt{\tan \left(x\right)} \setminus \mathrm{dx} \setminus =$

 = \ \ \sqrt{ 2 }/4 ( ln| { tan x - \sqrt{ 2 tan x } + 1 }/{ tan x + \sqrt{ 2 tan x } + 1 } | + 2 arctan( { \sqrt{ 2 tan x } }/{ 1 - tan x } ) ) + C. #

$\text{This is the desired result. } \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus q \quad \setminus \square$