Question

Question #e882e

Answer 1

See the answer below:

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!

Answer 2

`int \ \sqrt{ tan(x) } \ dx \ =`

# = \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. #

`"Please see solution below. Maybe simple ??!"`

Explanation:

`"We want to look at the integral:"`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad int \ sqrt{ tan(x) } \ dx.`

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

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

`"The one I found to be a bit better is:" \qquad \qquad u \ = \ sqrt{ 2 tan(x) }.`

`"So, let's go ahead with this substitution:" \qquad \ u \ = \ sqrt{ 2 tan(x) }.`

`"1) Let:" \qquad \qquad \qquad \qquad u \ = \ sqrt{ 2 tan(x) }. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \ \ \ (I)`

`"2) So:" \qquad \qquad \qquad \qquad du = { sec^2(x) } / { sqrt{ 2 tan(x) } } \ dx.`

`"3) Thus:" \qquad \qquad \qquad dx = { sqrt{ 2 tan(x) } } / { sec^2(x) } \ du.`

`"4) So, we begin (!!):"`

`\qquad \qquad \quad int \ sqrt{ tan(x) } \ dx`

`\qquad \qquad \qquad \qquad \qquad \quad = \ int \ sqrt{ tan(x) } cdot { sqrt{ 2 tan(x) } } / { sec^2(x) } \ du`

`\qquad \qquad \qquad \qquad \qquad \quad = \ sqrt{ 2 } int \ tan(x) / sec^2(x) \ du`

`\qquad \qquad \qquad \qquad \qquad \quad = \ sqrt{ 2 } \int \ tan(x) /{ 1 + tan^2(x) } \ du`

`\qquad \qquad \qquad \qquad \qquad \quad = \ sqrt{ 2 } \int \ { 1/2 u^2 } /{ 1 + 1/4 u^4 } \ du \qquad \qquad \qquad [ "as" \ \tan(x) \ = \ 1/2 u^2 ]`

`\qquad \qquad \qquad \qquad \qquad \quad = \ sqrt{ 2 } \int \ 4/4 cdot { 1/2 u^2 } /{ 1 + 1/4 u^4 } \ du`

`\qquad \qquad \qquad \qquad \qquad \quad = \ 2 sqrt{ 2 } \int \ u^2/{ 4 + u^4 } \ du.`

`\qquad \qquad :. \qquad \qquad int \ sqrt{ tan(x) } \ dx \ = \ 2 sqrt{ 2 } \int \ u^2/{ 4 + u^4 } \ du. \qquad \qquad \qquad \ (II)`

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

`"5) Expand integrand into partial fractions:"`

`\qquad "a) Factor denominator:"`

`\qquad \qquad \qquad \qquad \qquad \quad u^4 + 4 \ = \ u^4 + 4 u^2 + 4 - 4 u^2`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ ( u^2 + 2 )^2 - ( 2 u )^2`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ [ u^2 + 2 - 2 u ] cdot [ u^2 + 2 + 2 u ]`

`\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad = \ [ u^2 - 2 u + 2 ] cdot [ u^2 + 2 u + 2 ].`

`\qquad \qquad "Can tell the two factors on the RHS of the previous are"`
`"irreducible over" \ \ RR, "because their discriminants are both"`
`"negative. So:"`

`\qquad \qquad "Factorization of denominator, over" \ RR:`

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

`\qquad "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 }. #

`\qquad \qquad "Using the standard technique of clearing the"`
`"polynomial fractions, and equating coefficients of like powers"`
`"of" \ \ u \ \ "in the resulting equality of polynomials, and solving"`
`"the resulting linear system for" \ \ a, b, c, d \ \ "we find:"`

`\qquad \qquad \qquad a = 1/4 \qquad \qquad b = 0 \qquad \qquad c = -1/4 \qquad \qquad d = 0.`

`"[These values can easily be checked in the decomposition"`
`"equality above -- almost by eye alone !]"`

`\qquad \qquad "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 ) ). #

`"6) Completion of Integration:"`

`\qquad \qquad "Now we complete the integration, following from the"`
`"partial fraction decomposition, using the standard technique of"`
`"adjusting the polynomial fractions there, to set up a ln part,"`
`"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 } )#

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

`\qquad \qquad "Now by (II), and (III), we have, altogether:"`

`\ int \ sqrt{ tan(x) } \ dx \ = \ 2 \sqrt{ 2 } \int \ u^2/{ 4 + u^4 } \ 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 #

`\qquad \qquad \qquad \qquad \qquad [ \ "as, by (I):" \quad u = \sqrt{ 2 tan(x) } \qquad rArr \qquad ]`

# = \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. #

`\qquad \qquad "So, we now have here:"`

`\ int \ sqrt{ tan(x) } \ dx \ =`

# = \ \ \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. #

`"This is the desired result. " \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \ square`