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Integral of Trig Function, Exponent


Date: 8/6/96 at 18:47:16
From: Jose Eduardo Russo Sarmiento
Subject: Integral of Trig. Function, Exponent

Dr. Math:

I am teacher assistant in a calculus course, and one student asked 
for the primative [integral] of this function:

   f(x)=(tanx)^-1/2

Greetings from Venezuela, and I appreciate your help.


Date: 8/7/96 at 1:43:34
From: Doctor Pete
Subject: Re: Integral of Trig. Function, Exponent

According to Maple, the integral of your function is

       |                           |          Sqrt[Sin[2x]]
   Log |Sin[x]+Cos[x]+Sqrt[Sin[2x]]| - ArcTan -------------
       |                           |          Sin[x]-Cos[x] 
   -------------------------------------------------------- .
                           Sqrt[2]


Let Int[f(x),x] be the integral (or "primitive," as you say) of f(x) 
with respect to x.  Then we want to find

     S = Int[1/Sqrt[Tan[x]],x] .

Let z = Tan[x].  Then x = ArcTan[z], or dx = dz/(z^2+1).  Therefore 
our change of variables gives

     S = Int[1/(Sqrt[z]*(z^2+1)),z] .

Now, let u = Sqrt[z], or z = u^2, and dz = 2u du.  Then

     S = Int[2u/(u(u^4+1)),u]
       = 2*Int[1/(u^4+1),u] .

Here is the hard part.  We will factor u^4+1 into two irreducible 
quadratics and use partial fraction decomposition.  Notice that

     u^4+1 = (u^2+Sqrt[2]u+1)(u^2-Sqrt[2]+1) .

(You can check this by multiplying out the right hand side.)  Then 
suppose

                   1                      Au+B              Cu+D
   -------------------------------- = -------------- + --------------
   (u^2+Sqrt[2]u+1)(u^2-Sqrt[2]u+1)   u^2+Sqrt[2]u+1   u^2-Sqrt[2]u+1

for some unknown constants A, B, C, D.  Multiplying out both sides, we 
see that

     (Au+B)(u^2-Sqrt[2]u+1) + (Cu+D)(u^2+Sqrt[2]u+1) = 1 ,

or

     (A+C)u^3 + (-Sqrt[2]A+B+Sqrt[2]C+D)u^2 + (A-Sqrt[2]B+C+Sqrt[2]D) 
        + (B+D) = 1 .

Equating the left and right hand sides, we obtain the system

             A+                C          = 0            [Eq. 1]
     -Sqrt[2]A+       B+Sqrt[2]C+       D = 0            [Eq. 2]
             A-Sqrt[2]B+       C+Sqrt[2]D = 0            [Eq. 3]
                      B+                D = 1 .          [Eq. 4]

Substituting Eq. 4 into Eq. 2, we have -A+C = -1/Sqrt[2].  Using Eq. 
1, we find that A = 1/(2*Sqrt[2]), C = -A = -1/(2*Sqrt[2]).  
Substituting Eq. 1 into Eq. 3, we get -B+D = 0, and with Eq. 4, we 
find B = D = 1/2. Therefore,

     S = 2(Int[(u/(2*Sqrt[2])+1/2)/(u^2+Sqrt[2]u+1),u]
             -Int[(u/(2*Sqrt[2])-1/2)/(u^2-Sqrt[2]u+1),u]) .
       = 1/(2*Sqrt[2]) * (Int[(2u+2*Sqrt[2])/(u^2+Sqrt[2]u+1),u]
             -Int[(2u-2*Sqrt[2])/(u^2-Sqrt[2]u+1),u]) .

Here we notice that the derivative of u^2+Sqrt[2]u+1 = 2u+Sqrt[2], and 
the derivative of u^2-Sqrt[2]u+1 = 2u-Sqrt[2].  So

     S = 1/(2*Sqrt[2]) * (Log[u^2+Sqrt[2]u+1]-Log[u^2-Sqrt[2]u+1]
             + Int[Sqrt[2]/(u^2+Sqrt[2]u+1),u]
             + Int[Sqrt[2]/(u^2-Sqrt[2]u+1),u])
       = 1/(2*Sqrt[2]) * Log[(u^2+Sqrt[2]u+1)/(u^2-Sqrt[2]u+1)]
             + 1/2 * (Int[1/((u+1/Sqrt[2])^2 + (1/Sqrt[2])^2),u]
             + Int[1/((u-1/Sqrt[2])^2 + (1/Sqrt[2])^2),u])
       = 1/(2*Sqrt[2]) * Log[(u^2+Sqrt[2]u+1)/(u^2-Sqrt[2]u+1)]
             + 1/2 * (1/(1/Sqrt[2])) * (ArcTan[(u+1/Sqrt[2])/(1/
                 Sqrt[2])]
             + ArcTan[(u-1/Sqrt[2])/(1/Sqrt[2])])
       = 1/(2*Sqrt[2]) * Log[(u^2+Sqrt[2]u+1)/(u^2-Sqrt[2]u+1)]
             + 1/Sqrt[2] * (ArcTan[Sqrt[2]u+1] + ArcTan[Sqrt[2]u-1]) ,

and since ArcTan[x]+ArcTan[y] = ArcTan[(x+y)/(1-xy)],

     S = 1/(2*Sqrt[2]) * Log[(u^2+Sqrt[2]u+1)/(u^2-Sqrt[2]u+1)]
             + 1/Sqrt[2] * ArcTan[2*Sqrt[2]u/(2-2u^2)]
       = 1/(2*Sqrt[2]) * Log[(u^2+Sqrt[2]u+1)/(u^2-Sqrt[2]u+1)]
             + 1/Sqrt[2] * ArcTan[Sqrt[2]u/(1-u^2)] .
       = 1/(2*Sqrt[2]) * Log[(u^2+Sqrt[2]u+1)^2/(u^4+1)]
             + 1/Sqrt[2] * ArcTan[Sqrt[2]u/(1-u^2)] .

Substituting back (u = Sqrt[z]),

     S = 1/(2*Sqrt[2]) * Log[(z+Sqrt[2z]+1)^2/(z^2+1)]
             + 1/Sqrt[2] * ArcTan[Sqrt[2z]/(1-z)] .

Substituting back (z = Tan[x]),

     S = 1/(2*Sqrt[2]) * Log[(Tan[x]+Sqrt[2*Tan[x]]+1)^2/(Tan[x]^2+1)]
             + 1/Sqrt[2] * ArcTan[Sqrt[2*Tan[x]]/(1-Tan[x])]
       = 1/(2*Sqrt[2]) * Log[(Tan[x]+Sqrt[2*Tan[x]]+1)^2/Sec[x]^2]
             + 1/Sqrt[2] * ArcTan[Sqrt[2*Sin[x]/Cos[x]]/(1-Sin[x]/
               Cos[x])]
       = 1/(2*Sqrt[2]) * Log[(Cos[x]Tan[x]+Sqrt[2*Tan[x]Cos[x]^2]+
               Cos[x])^2]
             + 1/Sqrt[2] * ArcTan[Sqrt[2*Sin[x]Cos[x]]/(Cos[x]-
               Sin[x])]
       = 1/(2*Sqrt[2]) * 2 * Log[Sin[x]+Cos[x]+Sqrt[2*Sin[x]Cos[x]]]
             + 1/Sqrt[2] * ArcTan[Sqrt[Sin[2x]]/(Cos[x]-Sin[x])]
       = 1/Sqrt[2] * (Log[Sin[x]+Cos[x]+Sqrt[Sin[2x]]]
             + ArcTan[Sqrt[Sin[2x]]/(Cos[x]-Sin[x])] + C.

This is the same expression as I gave you before, except the sign on 
the ArcTan term is switched (this is because ArcTan[-x] = -ArcTan[x]).

I hope you understood my explanation; I have tried to write each step 
down, although I'm afraid the partial fraction decomposition part 
might be a little unclear.  Please let me know if you don't understand 
some step of the solution.

Doctor Pete,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   
    
Associated Topics:
College Calculus
High School Calculus

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