An Integration with SubstitutionDate: 06/06/98 at 22:05:28 From: Abecina Subject: Integrations Hello Dr Math, I am having a few problems with some integrations. Will you please help me? 1) Use the result of INT(0 to a) f(x) dx = INT (0 to a) f(a-x) dx to show that: x sin(x) pi^2 INT(0 to pi) ------------ dx = ---- 1 + cos^2(x) 4 2) If: x^n I(n) = INT(0 to 1) ------- dx x^2 + 1 1 i) Show that I(n) + I(n-2) = ----- n - 1 ii) Evaluate I(4) and I(5). Thank you. Date: 06/07/98 at 07:53:39 From: Doctor Anthony Subject: Re: Integrations Problem 1: Since sin(pi-x) = sin(x) and cos(pi-x) = -cos(x), we can use the above theorem to write the following equation: x sin(x) dx (pi-x) sin(x) dx I = INT ------------ = INT ---------------- 1 + cos^2(x) 1 + cos^2(x) Then adding these two values of I we get: pi sin(x)dx 2I = INT ------------ 1 + cos^2(x) Let cos(x) = u. Thus -sin(x) dx = du, and x = 0 implies u = 1 and x = pi implies u = -1. So: -pi du pi du 2I = INT(1 to -1) ------- = INT(-1 to 1) ------- 1 + u^2 1 + u^2 2I = pi tan^(-1)(u) from -1 to 1 = pi [pi/4 - (-pi/4)] = pi [pi/2] and so: I = pi^2/4 as required. Problem 2: (i) We have: x^n + x^(n-2) I(n) + I(n-2) = INT [-------------]dx x^2 + 1 x(n-2)(x^2 + 1) = INT[----------------]dx x^2 + 1 = INT[x^(n-2) dx] from 0 to 1 x^(n-1) = ------- from 0 to 1 n - 1 1 = ----- as required n - 1 (ii) Evaluate I(4) and I(5). I(4) + I(2) = 1/3 I(2) + I(0) = 1/1 and I(0) = INT[dx/(1+x^2)] = tan^(-1)(x) from 0 to 1 = pi/4 I(2) = 1 - pi/4 I(4) = 1/3 - (1 - pi/4) I(4) = pi/4 - 1 + 1/3 = pi/4 - 2/3 For I(5): I(5) + I(3) = 1/4 I(3) + I(1) = 1/2 and x dx I(1) = INT[-------] 1 + x^2 Let 1+x^2 = u. Thus 2x dx = du and x dx = (1/2) du. So: I(1) = (1/2)INT[du/u] where x=0 implies u=1 and x=1 implies u=2 = (1/2) ln(u) = (1/2)[ln(2)- ln(1)] = (1/2) ln(2) So I(5) = 1/4 - [1/2 - (1/2)ln(2)] = (1/2)ln(2) - 1/2 + 1/4 = (1/2)ln(2) - 1/4 -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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