Method of Euler Substitution

The Method of Euler Substitution PhL 11.1.16 Warning: The first 10 sections are valid. In the last 4 sections I was just flailing away trying to find the sin-1 form of my integral. Sometimes bad paths lead to lots of work and no payoff. This form has since been easily derived in my newer doc. I may publish this at some point, would be a very short doc. Just to save someone else from having to do the work. 1. Introduction 1 2. Comments about 2 3. Naive method, does not work 3 4. Improved method A 4 5. Improved method B 7 6. Improved method C 10 7. Improved method D (my own, no good) 11 8. Summary of the three methods that are useful (these are Euler's three) 12 9. Application of Method A 13 10. Redo the same Application using Method B 18 11. Try to convert this to sin-1 form (did not work) 26 12. Go back to the log form which is perhaps the simplest and most flexible, 28 13. Try adding a factor f(a,b,c) 30 14. Start once again 31 1. Introduction The problem is how to do integrals of this form (see GR7 page 92) ∫dx R(x,) What does this notation mean? In GR7 this means a "rational function". Usually a rational function is a ratio of polynomials. But what does R(a,b) mean? It is never defined in GR7 as far as I know (nor on the web). I think what it means is that the integrand is a ratio of polynomials in the two variables a and b. Examples R(x,) = typical The numerator and denominator are separately polynomials in these two variables. Now since you can treat the terms separately, the numerator can be taken to be a monomial, so then R(x,) = . As I browse through the GR7 integrals starting at page 92 they seem to have this form. The main issue is this: How to you get rid of the ugly radical . Warning: Things are then confused when the same symbol R is redefined to mean R ≡ a + bx + cx2 Almost all integrals of interest, however, seem to be of the form F(x,) = xm ()n where m and n are arbitrary integers of either sign. So this would be a special case of the above more general form. 2. Comments about 1. The a,b,c notation I (and GR) have chosen matches high-school usage. Some authors (Schuam) have a↔c so beware. I assume that a,b,c are all real, at least as an initial position. 2. If c > 0, we have no ambiguity in writing = = If c < 0, we have to decide on a phase for = eiπ/2 probably is what I would choose. 3. Now write R/c = x2 + (b/c)x + (a/c) = (x-α+)(x-α-) where α± are the roots. Then have = = R = a + bx + cx2 = c(x-α+)(x-α-) 4. Values α± are also the roots of a + bx + cx2 regardless of the sign of c. 5. These roots of x2 + (b/c)x + (a/c) = 0 are given by the quadratic formula, α± = [ -b/c ± ] / 2 = [ -b/c ± (1/c) ] / 2 = [ -b ± ] / 2c where b2-4ac is called "the discriminant". 5a. If it happens that b2 = 4ac, then we get α± = -b/(2c) and then R = c(x-α+)2 = c(x + [b/2c])2 and = (x + [b/2c] ). This is then for Δ = 0 as defined in the next item. 6. GR7 make this definition -Δ ≡ b2 - 4ac 7. If b2 - 4ac > 0 then the roots are real. This means that the function f(x) = a + bx + cx2 has two intersections with the x axis. If c > 0, the parabola cups up and intersects the x axis at the two roots. Therefore to the right of the upper root, we have a + bx + cx2 and so is real and well defined. 8. If c < 0, the parabola is cupping down. If b2 - 4ac > 0 the roots are real as before, but now is well defined and real only between the two roots. 9. When talking about integrals involving , I feel it is best to start with an integral over a range of x where a + bx + cx2 is positive, so then is real. Thus we have Case 1: c > 0, b2- 4ac > 0 (real roots), cups up, x > α+ or x < α- to have real Case 2: c > 0, b2- 4ac < 0 (imag roots), cups up, all values of x give real Δ > 0 Case 3: c < 0, b2- 4ac > 0 (real roots), cups down, must have α- < x < α+ to have real Case 4: c < 0, b2- 4ac < 0 (imag roots), cups down, no value of x gives real It seems to me that Case 2 is the safest to deal with, since is real for any value of x, so when you integrate you don't have to worry about what values x takes in the integration. Fact: I will assume Case 2 from now on. This means c > 0 and Δ > 0 in language of GR7. 3. Naive method, does not work Ones first thought for "change of variables" would be to just set = t where t is a new variable. We would then be talking about R(x,) → R(x(t),t) But we want x(t) to be simple! With the above method we get a + bx + x2 = t2 x2+bx+(a-t2) = 0 and then x(t) is a quadratic formula x(t) = [-1 ± ]/2 and this then causes the xm factors to be a mess still involving radicals. So this choice is no go. That is to say, when we change variables from x to t, we don't want to have radicals in the resulting functions of t. 4. Improved method A Suppose instead you tried this = = t + x // = = t - x With Case 2 above, is real, , x is real and t is real, so there are no issues with ranges of x or t. Now we get R = a + bx + cx2 = t2 + cx2 + 2t x // R = a + bx + cx2 = t2 + cx2 - 2t x Now you have arranged to have two terms cancel out with this result a + bx = t2 + 2t x (b - 2t )x = (t2-a) x(t) = (t2-a)/(b - 2t ) // x(t) = (t2-a)/(b + 2t ) Maple verify: Maple continues which says = (bt-t2-a)/(b - 2t) // = (bt+t2+a)/(b + 2t) You then end up with R[x,] → R[(t2-a)/(b - 2t), (bt-t2-a)/(b - 2t)] and your result then has no ugly radicals at least, though it is still a mess. BUT, could an ugly radical reappear when you compute the connection between dx and dt? x(t) = (t2-a)(b - 2t )-1 dx/dt = (t2-a)(-1) (b - 2t )-2 (- 2) + (2t) (b - 2t )-1 or dx/dt = (2)(t2-a) (b - 2t )-2 + 2t (b - 2t )-1 = 2 (b - 2t )-2 [ ()(t2-a) + t(b - 2t ) ] = 2 (b - 2t )-2 [t2 - a + tb - 2t2 ] = 2 (b - 2t )-2 (bt - a -t2 ) Maple verify: OK so then x(t) = (t2-a)(b - 2t )-1 = (bt-t2-a)/(b - 2t) dx = 2 (b - 2t )-2 (bt - a -t2 ) dt No, no bad radicals have appeared here! That is because there were no radicals in x(t). So we then have !Syntax Error, Idx R(x,) = !Syntax Error, Idt * 2 (b - 2t )-2 (bt - a -t2 ) * R[ (t2-a)/(b - 2t), (bt-a-t2)/(b - 2t) ] It ain't purdy but at least there are no nasty radicals. But I also need to know t(x) so go back, a + bx = t2 + 2t x t2 + [2 x] t + [-a-bx] = 0 t = [- 2 x ± ]/2 = - x ± so then the integration endpoint is t±(x) = - x ± // t±(x) = + x ± and we eventually have to worry about the sign. Both these functions t±(x) are real with our Case 2 conditions. Both these t functions satisfy [ ]2 = [ t + x]2 . But we want t to satisfy the equation = t + x ! Which sign works? = t+ + x = [- x + ] + x = // = t+ - x = [+ x + ] - x = Therefore the only correct solution is t+ which we will call t. Then t(x) = - x + . // t(x) = + x + Comment: Think of this more traditional example !Syntax Error, If(sinθ)dθ u(θ) = sinθ du = cosθdθ dθ = (1-u2)-1/2du !Syntax Error, If(sinθ)dθ = !Syntax Error, If(u) (1-u2)-1/2du Example: f(x) = x so have !Syntax Error, Isinθ dθ = !Syntax Error, I u (1-u2)-1/2 du = - |u=sin(θ) = - cosθ Summary of Method A: = = t + x x(t) = (t2-a)(b - 2t )-1 = (bt-t2-a)/(b - 2t) dx = 2 (b - 2t )-2 (bt - a -t2 ) dt t(x) = - x + !Syntax Error, Idx R(x,) = !Syntax Error, Idt * 2 (b - 2t )-2 (bt - a -t2 ) * R[ (t2-a)/(b - 2t), (bt-a-t2)/(b - 2t) ] Question: How does Method A change if you take a different sign and say = t - x ? I have reviewed lines above, and the answer is that you just take → - in ALL results. Of course this implies that c → c with no change. So the above would be + = = t - x x(t) = (t2-a)(b + 2t )-1 = (bt+t2+a)/(b + 2t) dx = 2 (b + 2t )-2 (bt + a +t2 ) dt t(x) = + x + !Syntax Error, Idx R(x,) = !Syntax Error, Idt * 2 (b + 2t )-2 (bt + a +t2 ) * R[ (t2-a)/(b + 2t), (bt+a+t2)/(b + 2t) ] Perhaps a little nicer in the blue form because the final result has all + signs. 5. Improved method B Suppose instead you tried this = xt + Now we get a + bx + cx2 = x2t2+ a + 2 xt Now you have arranged to have two terms cancel out with this result bx + cx2 = x2t2+ 2 xt or b + cx = xt2 + 2 t x(c-t2) = 2 t - b x(t) = (2 t - b)/(c-t2) Maple verify: Maple has shown that x(t) = (2 t - b)/(c-t2) = (t2-tb+c)/(c-t2) dx = 2 (t2-tb+c)(t2-c)-2 dt Then: !Syntax Error, Idx R(x,) = !Syntax Error, Idt * 2 (t2-tb+c)(t2-c)-2 * R (2t - b)/(c-t2), (t2-tb+c)/(c-t2)] We then also need t(x): b + cx = xt2 + 2 t // b + cx = xt2 - 2 t xt2 + 2 t + [-b-cx] = 0 t± = [-2± ] /2x = [ - ± ] /x = [ - ± ] /x // t± = [ + ± ] /x We now confirm which is the correct solution going back to = xt+ + = x { [ - + ] /x} + } = { [ - + ]} + = Repeat this in blue: = xt+ - = x { [ + + ] /x} - } = { [ + + ]} - = Summary: = xt + x(t) = (2 t - b)/(c-t2) t(x) = [ - + ] /x = (t2-tb+c)/(c-t2) dx = 2 (t2-tb+c)(t2-c)-2 dt !Syntax Error, Idx R(x,) = !Syntax Error, Idt * 2 (t2-tb+c)(t2-c)-2 * R (2t - b)/(c-t2), (t2-tb+c)/(c-t2)] I think you could replace → - (and a→a) to get a "blue form" for the above equations. 6. Improved method C Write a + bx + cx2 = c [ x2 + (b/c)x+ (a/c) ] = c (x-α)(x-β) where α and β are the roots of a + bx + cx2 . Then try this: = (x-α)t Then a + bx + cx2 = (x-α)2t2 or c (x-α)(x-β) = (x-α)2t2 or c(x-β) = (x-α)t2 where once again we have cancellation on the two sides. Then cx - cβ = xt2-αt2 or x(c - t2) = cβ-αt2 x(t) = ( cβ-αt2)/(c-t2) So we have the following x(t) = (cβ-αt2)/(c-t2) = c(β-α)t/ (c-t2) dx = [ 2c(β-α)t / (c-t2)2] dt Then, !Syntax Error, Idx R(x,) = !Syntax Error, Idt * [ 2c(β-α)t / (c-t2)2] * R[ (cβ-αt2)/(c-t2) , c(β-α)t/ (c-t2)] This one is perhaps more complicated because you have to compute the two roots α and β, but you can see that it gives a result with no radicals, as in the previous two cases. In the case of my Example below this will give !Syntax Error, Idx/ = !Syntax Error, Idt * [ 2c(β-α)t / (c-t2)2] * [ c(β-α)t/ (c-t2)]-1 = !Syntax Error, Idt * [ 2 / (c-t2)] = same result as Method B shown below. 7. Improved method D (my own, no good) Suppose instead you tried this R = = t + Now we get R = a + bx + cx2 = t2 + bx + 2t Again we get two terms cancelling, leaving a +cx2 = t2 +2t But this method does not give a simple expression for x(t), so it is a no go. 8. Summary of the three methods that are useful (these are Euler's three) Method A: = = t + x x(t) = (t2-a)(b - 2t )-1 t(x) = - x + = (bt-t2-a)/(b - 2t) dx = 2 (b - 2t )-2 (bt - a -t2 ) dt !Syntax Error, Idx R(x,) = !Syntax Error, Idt * 2 (b - 2t )-2 (bt - a -t2 ) * R[ (t2-a)/(b - 2t), (bt-a-t2)/(b - 2t) ] Method B: = xt + x(t) = (2 t - b)/(c-t2) t(x) = [ - + ] /x = (t2-tb+c)/(c-t2) dx = 2 (t2-tb+c)(t2-c)-2 dt = !Syntax Error, Idx R(x,) = !Syntax Error, Idt * 2 (t2-tb+c)(t2-c)-2 * R (2t - b)/(c-t2), (t2-tb+c)/(c-t2)] Method C: where α,β are the roots of a + bx + cx2 = 0 " = (x-α)t x(t) = (cβ-αt2)/(c-t2) t(x) = = c(β-α)t/ (c-t2) dx = [ 2c(β-α)t / (c-t2)2] dt !Syntax Error, Idx R(x,) = !Syntax Error, Idt * [ 2c(β-α)t /(c-t2)2] * R[ (cβ-αt2)/(c-t2) , c(β-α)t/ (c-t2)] 9. Application of Method A First I quote the Method A results x(t) = (t2-a)(b - 2t )-1 t(x) = - x + = (bt-t2-a)/(b - 2t) dx = 2 (b - 2t )-2 (bt - a -t2 ) dt !Syntax Error, Idx R(x,) = !Syntax Error, Idt * 2 (b - 2t )-2 (bt - a -t2 ) * R[ (t2-a)/(b - 2t), (bt-a-t2)/(b - 2t) ] Now the application is this: R(x,) = 1/ so !Syntax Error, Idx/ = !Syntax Error, Idt * 2 (b - 2t )-2 (bt - a -t2 ) * 1/ [ (bt-t2-a)/(b - 2t) ] = !Syntax Error, Idt * 2 (b - 2t )-2 (bt - a -t2 ) * (b - 2t) / (bt-t2-a) = !Syntax Error, Idt * 2 (b - 2t )-1 (bt - a -t2 ) * 1 / (bt-t2-a) = !Syntax Error, Idt * 2 (b - 2t )-1 * 1 / 1 = 2!Syntax Error, Idt / (b - 2t ) = 2/(2)!Syntax Error, Idt / (b/(2) - t ) = - (1/) !Syntax Error, Idt / ( t - r) r ≡ b/(2) = - (1/) ln(t-r)|t(x) = - (1/) ln(t(x)- r) . // algebra seems OK But recall that t(x) = - x + = - x + so we then have = - (1/) ln[- x + - b/(2)] = - (1/) ln[- x - b/(2) + ] Now think of this as ln[ ] . You are allowed to multiply top and bottom by whatever you like that does not depend on x. Then you can throw away the bottom since it is just a constant indep of x. So mult top and bottom by + 2 to get = - (1/) ln [ - 2cx - b + 2] Again, I have done this in Case 2 where c > 0, -Δ = b2- 4ac < 0 so Δ > 0 There are two official GR7 answers for this case The result I got is similar to the first result here, but it is NOT THE SAME even adding constants. I seem to have multiple sign errors! The denominator factor is irrelevant since it just represents a constant, so here is the GR7 result compared with my result GR7(x) = + (1/) ln [ +2cx + b + 2] PL(x) = - (1/) ln [ - 2cx - b + 2] So I claim to have shown that !Syntax Error, Idx/ = PL(x) + constant A check would be to show that dPL(x)/dx = 1/ ? OK, dPL(x)/dx = - (1/) dx ln [ - 2cx - b + 2] = - (1/) [ - 2cx - b + 2]-1 (-2c + 2 dx ) Now dx = (1/2)(1/)(b + 2cx) So then, dPL(x)/dx = - (1/) [ - 2cx - b + 2]-1 [-2c + 2(1/2)(1/)(b + 2cx) ] = - (1/) [ - 2cx - b + 2]-1 [-2c + (b + 2cx) ] / = (1/) [ - 2cx - b + 2]-1 [2c - (b + 2cx) ] / = (1/) [ - 2cx - b + 2]-1 [2c - b - 2cx) ] / = (1/) [ - 2cx - b + 2]-1 [2 - b - 2cx) ] / = [ - 2cx - b + 2]-1 [2 - b - 2cx) ] / = [ - 2cx - b + 2]-1 [ - 2cx - b + 2 ) ] / = 1/ This would seem to say that my result is correct. Now let's try their result: dGR7(x)/dx = (1/) dx ln [ +2cx + b + 2] = (1/) [ +2cx + b + 2]-1 ( 2c + 2 dx ) = (1/) [ +2cx + b + 2]-1 ( 2c + 2 [(1/2)(1/)(b + 2cx)] ) = (1/) [ +2cx + b + 2]-1 ( 2c + (1/)(b + 2cx) ) = (1/) [ +2cx + b + 2]-1 ( 2c + (b + 2cx) )/ = (1/) [ +2cx + b + 2]-1 ( 2c + b + 2cx) )/ = (1/) [ +2cx + b + 2]-1 ( 2 + b + 2cx) )/ = [ +2cx + b + 2]-1 ( 2 + b + 2cx) )/ = 1/ So both the PL and GR7 solutions are valid! How can that be? GR7(x) = + (1/) ln [ +2cx + b + 2] PL(x) = - (1/) ln [ - 2cx - b + 2] = (1/) ln [ (- 2cx - b + 2)-1] = (1/) ln [ ] Now mult top and bottom by (- 2cx - b - 2), certainly allowed to do that = (1/) ln [ ] Now evaluate the bottom: (- 2cx - b + 2)(- 2cx - b- 2) = (2cx + b - 2)(2cx + b+ 2) = (2cx + b)2 - 4cR = (2cx + b)2 - 4c(a+bx+cx2) = 4c2x2 + b2 + 4cxb - 4ca - 4cbx - 4c2x2 = b2 - 4ca Thus I have shown that PL(x) = (1/) ln [ ] = (1/) ln [ ] = (1/) ln [ ] // dropping constant term = (1/) ln [ ] // dropping constant term = (1/) ln[ +2cx + b + 2] = GR7(x) So finally I have shown that my result agrees with the GR7 result. Question: What happens to this example if we take the blue form? My black result was PL(x) = - (1/) ln [ - 2cx - b + 2] so following the simple rule → - this becomes PL(x) = + (1/) ln [ - 2cx - b - 2] Now use the fact that ln[-s] = ln(eiπs) = iπ + ln(x) and throw out the constant to get PL(x) = + (1/) ln [ 2cx + b + 2] Recall from above GR7(x) = + (1/) ln [ +2cx + b + 2] so using the blue method gets you the GR7(x) result, good! Here is Maple's version of the integral: which I can write as (inside log, to top and bottom times 2) and then drop bottom constant : (1/) ln ( b + 2cx + 2) and this agrees with the GF7 function stated above. 10. Redo the same Application using Method B First pull in the Method B summary, = = xt + // = xt - x(t) = (2 t - b)/(c-t2) t(x) = [ - + ] /x = (-+ )/x = (t2-tb+c)/(c-t2) dx = 2 (t2-tb+c)(t2-c)-2 dt !Syntax Error, Idx R(x,) = !Syntax Error, Idt * 2 (t2-tb+c)(t2-c)-2 * R (2t - b)/(c-t2), (t2-tb+c)/(c-t2)] Now let R(x,) = 1/ so the above becomes !Syntax Error, Idx/ = !Syntax Error, Idt * 2 (t2-tb+c)(t2-c)-2 * 1/ (t2-tb+c)/(c-t2) = !Syntax Error, Idt * 2 (t2-tb+c)(t2-c)-2 * (c-t2)/ (t2-tb+c) = - !Syntax Error, Idt * 2 (t2-c)-1 = - 2!Syntax Error, Idt /(t2- c) I can have Maple do this integral to get Continuing then from above !Syntax Error, Idx/ = - 2!Syntax Error, Idt /(t2- c) = + (2/) tanh-1(t/)|t=t(x) = (2/)tanh-1 [ (-+ )/(x)] = (2/) tanh-1 [ ] I guess I have assumed that a > 0 and c > 0 just by writing and . GR7 does not give a result in this form. Let's first just check doing the derivative dx { (2/) tanh-1 [ ] } = (2/) dx { tanh-1 [ ] } = (2/) dx { tanh-1 u(x) } // recall dx = (1/2)(1/)(b + 2cx) = (2/) (1-u2)-1dx [ ] = (2/) (1-u2)-1{ x dx - (-+ )} / x2c = (2/) (1-u2)-1{ x (1/2)(1/)(b + 2cx) - (-+ )} / x2c = (2/) (1-u2)-1{ x (1/2)(1/)(b + 2cx) - (-+ )} / x2c = 2 (1-u2)-1 { x (1/2)(1/)(b + 2cx) - (-+ )} / x2c Meanwhile: u = 1 - u2 = 1 - [ ]2 = [ x2c - (-+ )2 ] / x2c = [ x2c - (a - 2+ R) ] / x2c = [ x2c -a + 2- R) ] / x2c Now we end up with dx(sol) = 2 { x2c / [ x2c -a + 2- R) ]} * { x (1/2)(1/)(b + 2cx) - (-+ )} / x2c = 2 {1/ [ x2c -a + 2- R) ]} * { x (1/2)(1/)(b + 2cx) + - )} = 2 {1/ [ x2c -a + 2- R) ]} * { x (1/2)(b + 2cx) + - R )} * (1/) = 1 { [ x (b + 2cx) + 2- 2R )] / [ x2c -a + 2- R) ]} * (1/) = N/D * (1/) To make things work, I would need to have N = D. But, N = x (b + 2cx) + 2- 2R ) = xb + 2xc2 + 2 - R - a - bx - cx2 = xc2 + 2 - R - a Meanwhile from above we have D = x2c -a + 2- R Wow, so I have N = D, and the result is verified! So I have a "new form" of the integral, I ≡ !Syntax Error, Idx/ = (2/) tanh-1 [ ] which I have directly verified by differentiation. Question: Why would this still be valid if → - doing the "blue" pathway. Maybe it would help to write the above in its log form using tanh-1u = (1/2) ln[(1+u)/(1-u)] u = (- + )/(x) = (1/2) ln [ ] = (1/2) ln [ ] = (1/2) ln [ ] Now make the change → - to get [blue] = (1/2) ln [ ] = To show these are equivalent, I have to show that ln [ ] - ln [ ] = constant independent of x which says ln [ * ] = const ?? = ln [num/den] num = (x- )2 - R = x2c + a - 2x - a - bx - cx2 = - 2x - bx den = (x+ )2 - R = x2c + a + 2x - a - bx - cx2 = +2x - bx Note that num/den = ( - 2x - bx)/(2x - bx) = ( - 2 - b)/(2 - b) = - 4ac + b2 Thus we have ln [num/den] = ln [- 4ac + b2] and this IS in fact a constant independent of x. Fact: If you do the blue path and take → - , you get another valid solution! I ≡ !Syntax Error, Idx/ = (2/) tanh-1 [ ] + K I ≡ !Syntax Error, Idx/ = (2/) tanh-1 [ ] + K' Here is Maple confirming that these are both valid forms: How to I express this in terms of other hyperbolics? Start this way with the second solution above, (/2) I = tanh-1 [ ] OK tanh { (/2) I } = OK tanh2 { (/2) I } = []2 Then sech2 { (/2) I } = 1 - tanh2 { (/2) I } = 1 - []2 = [ cx2 - ( + )2 ] / (x)2 sech{ (/2) I } = [ cx2 - ( + )2 ]1/2 / (x) (/2) I = sech-1 [ cx2 - ( + )2 ]1/2 / (x) I = (2/ ) sech-1 { [ cx2 - ( + )2 ]1/2 / (x) } = (2/ ) sech-1 { [ -2a - bx - 2 ]1/2 / (x) } = (2/ ) sech-1 ( ) = (2/ ) sech-1 ( ) I have not verified this result by differentiation, it is just a waypoint. But since later result is verified, this must be valid as well. Now cosh2 {(/2) I} = { sech2 {(/2) I}-1 = (x)2 / [ cx2 - ( + )2 ] cosh{(/2) I} = (x) / [ cx2 - ( + )2 ]1/2 (/2) I = cosh-1 { (x) / [ cx2 - ( + )2 ]1/2 } I = (2/)cosh-1 { (x) / [ cx2 - ( + )2 ]1/2 } = (2/)cosh-1 { (x) / [ cx2 - a - 2 - R ]1/2 } = (2/)cosh-1 { (x) / [ -2a - bx - 2 ]1/2 } I enter the first form into Maple as J3 because I want Maple to do what seems to me is a messy derivative, I replace R = a + bx + cx2 and then I differentiate to get This is a big mess, and Maple refuses to simplify the denominator in the obvious manner that a person would do, and I don't know how to beat this problem. So my next step is to break Q = Qnum/Qden and replace by "r". This then gives for the denominator I manually simplify that to get Qden = (2a+2+bx) F1/2 [x2c - (-2a-2 -bx) ]1/2 F-1/2 = (2a+2+bx) [x2c+2a+2+bx ]1/2 In the square root I replace x2c = r - a - bx to get [x2c+2a+2+bx ]1/2 = [r - a - bx +2a+2+bx ]1/2 = [r + a +2 ]1/2 = [(+)2]1/2a = (+) which is a crucial step. I then have this greatly simplified denominator, Qden = (2a+2+bx) (+) I then manually enter this into Maple to override the previous expression for Qden., I want now to show that Qnum/Qden = 1/ and then I will have verified the above integral I. So define δ = Qden/ - Qnum and the goal then is to show that δ = 0: This took me about 1.5 days to get right! So at this point I know have several forms for my integral I ≡ !Syntax Error, Idx/ Here is what I have so far, where each form has been verified by differentiation. The first two come using Method A above, I = - (1/) ln [ - 2cx - b + 2] + const // this is PL(x) from above (1) I = (1/) ln [ + 2cx + b + 2] + const // this is GL7(x) from above (Maple too) (2) Form (2) agrees with GR7 and with Maple's evaluation of this integral. I showed that form (1) is valid if form (2) is valid. Form (2) also appears in Schaum p 72 top. If b2 = 4ac, then item 5a. above tells us that = (x + [b/2c] ). In this case we get 2cx + b + 2 = 2cx + b + 2 (x + [b/2c] ) = 2cx + b + 2cx +b = 4cx + 2b Then form (2) says I(Δ=0) = (1/) ln [4cx + 2b] = (1/) ln [2cx + b] in agreement with GR7. Alternatively, we can do the integral directly for Δ = 0 to get, !Syntax Error, Idx/ = !Syntax Error, Idx = (1/) !Syntax Error, Idx = (1/) ln (x + [b/2c]) = (1/) ln ( ) = (1/) ln (2cx +b) From Method B I get two more diff-verified forms I ≡ !Syntax Error, Idx/ = (2/) tanh-1 [ ] + const (3) I ≡ !Syntax Error, Idx/ = (2/) tanh-1 [ ] + const (4) I then convert this to some other forms. First I get, I = (2/) sech-1 ( ) + const (5) = (2/) sech-1 ( ) + const (6) Then comes the most recent result which is diff-verified, I = (2/)cosh-1( ) + const 11. Try to convert this to sin-1 form (did not work) cosh {(/2) I} = = cosh {(-i/2) I} = i = my choice of sign = cosh {(i/2) I} = cos {(/2) I} So at this point I have cos {(/2) I} = Now use a half-angle formula to write cos(2x) = 2 cos2(x) - 1 or cos(x) = 2 cos2(x/2) - 1 Then have cos {() I} = 2 []2 - 1 = 2 - 1 = [ 2cx2 - cx2 + ( + )2 ] / [cx2 - ( + )2] = [ cx2 + ( + )2 ] / [cx2 - ( + )2] So my latest result is then this I = (1/) cos-1 () Now since cos-1 = π/2 - sin-1, this implies that I = - (1/) sin-1 () + K1 How do you get from this result to this GR7 result + K2 You would have to show that - (1/) sin-1 () + K1 = - (1/) sin-1() + K2 or - (1/) sin-1 () = - (1/) sin-1() + K3 or sin-1 () = sin-1() - K3 That looks like a lot of work to prove or disprove. You might have hoped that = = or N = D (2cx+b) or N2(b2-4ac) = D2(2cx+b)2 or T1 = T2 Try this in Maple. // It does not work, they are not equal. 12. Go back to the log form which is perhaps the simplest and most flexible, I = (1/) ln [ 2cx + b + 2] // this is GL7(x) I am allowed to replace this with I = (1/) ln [ (2cx + b + 2) f(a,b,c)] where f is an arbitrary function! Can I get the argument of ln to have this form y + = (2cx + b + 2) = (2cx + b + 2)- y (y2+1) = (2cx + b + 2- y)2f2 = y2 - 2(2cx + b + 2)y + (2cx + b + 2)2 so 1 = - 2(2cx + b + 2)y+ (2cx + b + 2)2 2(2cx + b + 2)y = (2cx + b + 2)2 - 1 y = OK, then according to Schaum p 29 I can claim that I = (1/) ln [ 2cx + b + 2] = (1/) ln [y + ] = (1/) sinh-1 y = (1/) sinh-1 [] That was pretty fast. Now Schaum p 72 claims this should be = (1/) sinh-1 I don't see how that can happen even accounting for a generic constant. You would need to have this be true, (1/) sinh-1 [] = (1/) sinh-1 + f(a,b,c) or sinh-1 [] - sinh-1 = g(a,b,c) or sh-1X -sh-1Y = g or α - β = g shα = X shβ = Y OK, apply sh to get sh(α-β) = sh(g) shα chβ - chαshβ = sh(g) shα - shβ = sh(g) X - Y = sh(g) Now we have X = Y = I can ask Maple now whether it is possible that X - Y = a function not of x ! It seems that this has to be true for both forms for I to be correct. But this is too much mess for Maple, needs to be spoon fed. So let A = (2cx+b+2) X = (A2-1)/2A 1+X2 = 1 + (A2-1)/2A = [2A+A2-1]/(2A) B = Y = (2cx+b)/B 1+Y2 = 1 + (2cx+b)/B = [B+2cx+b]/B Then = / = / X - Y = (A2-1)/2A* / - (2cx+b)/B * / = (A2-1)(2A)-3/2 - (2cx+b)B-1/2 (2A)3/2 B1/2 [ X - Y] = (A2-1)B1/2 - (2cx+b)(2A)3/2 This is something Maple could evaluate. I could then want to show that when evaluated, you get (A2-1)B1/2 - (2cx+b)(2A)3/2 = (2A)3/2 B1/2 sh(g) where g does not depend on x. I would say this has a 1% chance at most. // Did not work. STOP. I need a more direct way to get the sh-1 form of the integral. sh (sh-1X -sh-1Y ) = shg or sh (sh-1X)ch(sh-1Y) - ch(sh-1X) sh(sh-1Y) = shg or X ch(sh-1Y) - ch(sh-1X) Y = shg 13. Try adding a factor f(a,b,c) Suppose I insert my arbitrary function by replacing (2cx + b + 2) → (2cx + b + 2) f. Then = (1/) sinh-1 [] Can I find a function f(a,b,c) such that = Define g ≡ (2cx + b + 2) . Then we have (g2f2-1)/2gf = (g2f2-1) = 2gf g2f2 - 2gf - 1 = 0 or (gf)2 + B(gf) - 1 = 0 B = - This says fg = [ -B ± ]/2 Now take a look see: B2+ 4 = (2cx+b)2/(4ac-b2) + 4(4ac-b2)/(4ac-b2) = [ (2cx+b)2 + 4(4ac-b2)] / (4ac-b2) = [ 4c2x2 + b2 + 4cxb +16ac - 4b2] / (4ac-b2) = [ 4c2x2 -3b2 + 4cxb +16ac ] / (4ac-b2) Then you get this ugly result fg = [ ± / ] /2 or 2fg = [ 2cx+b ± ] / g ≡ (2cx + b + 2) This is leading nowhere. Status: I am unable to derive the sin-1 or sinh-1 forms. 14. Start once again Assume the exact form given in GR7 so that I = (1/) sh-1(y) y = [2cx+b+2] / Then another form for the solution should be I = (1/) ln [ y + ] To help Maple, define A = (2cx+b+2) B = y = A/B Then another form for I should be I = (1/) ln [ (A/B) + ] = (1/) ln [ (A/B) + (1/B) ] = (1/) ln [A + ] // dropping ln(B) What we really want is to find that A + = f(a,b,c) A and then you end up with the GR7 ln form for I. So what you need is = g A (A2+B2) = g2A2 A = (2cx+b+2) This seems very unlikely. Maple says, Do it by hand (2cx+b+2)2 + (4ac-b2) = g2 (2cx+b+2)2 ?? This DOES NOT FLY, so I don't thing to two forms are linked in this manner.