Integral of Trig Function, ExponentDate: 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/ |
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