Question #e882e

2 Answers

See the answer below:
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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 \ \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