Integrate using trigonometric substitution !!?
it can go w/o subsitution, integrate dx/sqrt(4x^2-49) 1/2 integrate dx/sqrt. [x^2-(7/2)^2] 1/2 *1/2*2/7 ln(2x-7)/(2x+7) 1/14 ln (2x-7)/(2x+7)
Using trigonometric substitution. solve the integral of x^2/(the square root of x^2 -4) dx?
∫ [x² /√(x² - 4)] dx = let x = 2secθ secθ = x/2 dx = 2tanθ secθ dθ substitute, yielding: ∫ [x² /√(x² - 4)] dx = ∫ {(2secθ)² /√[(2secθ)² - 4]} 2tanθ secθ dθ = ∫ [4sec²θ /√(4sec²θ - 4)] 2tanθ secθ dθ = factor 4 out of the radical: ∫ {4sec²θ /√[4 (sec²θ - 1)]} 2tanθ secθ dθ = replace (sec²θ - 1) with tan²θ: ∫ {4sec²θ /[2√(tan²θ)]} 2tanθ secθ dθ = ∫ [4sec²θ /(2tanθ)] 2tanθ secθ dθ = simplifying, you have: 4 ∫ sec²θ secθ dθ = let: sec²θ dθ = dv → tanθ = v secθ = u → tanθ secθ dθ = du integrating by parts, you get: ∫ u dv = v u - ∫ v du 4 ∫ sec²θ secθ dθ = 4 (tanθ secθ - ∫ tanθ tanθ secθ dθ) → 4 ∫ sec³θ dθ = 4tanθ secθ - 4 ∫ tan²θ secθ dθ → replace tan²θ with (sec²θ - 1): 4 ∫ sec³θ dθ = 4tanθ secθ - 4 ∫ (sec²θ - 1) secθ dθ → expand it into: 4 ∫ sec³θ dθ = 4tanθ secθ - 4 ∫ (sec³θ - secθ) dθ → split it into: 4 ∫ sec³θ dθ = 4tanθ secθ - 4 ∫ sec³θ dθ + 4 ∫ secθ dθ → collect ∫ sec³θ dθ on the left side: 4 ∫ sec³θ dθ + 4 ∫ sec³θ dθ = 4tanθ secθ + 4 ∫ secθ dθ → (2)4 ∫ sec³θ dθ = 4tanθ secθ + 4 ∫ secθ dθ → 4 ∫ sec³θ dθ = (1/2) (4tanθ secθ + 4 ∫ secθ dθ) = 2tanθ secθ + 2 ∫ secθ dθ = multiply the remaining integrand by (tanθ + secθ)/(secθ + tanθ) (= 1): 2tanθ secθ + 2 ∫ (tanθ + secθ) secθ dθ /(secθ + tanθ) = expand the top: 2tanθ secθ + 2 ∫ (tanθ secθ + sec²θ) dθ /(secθ + tanθ) = note that the top is the derivative of the bottom: 2tanθ secθ + 2 ∫ d(secθ + tanθ) /(secθ + tanθ) = 2tanθ secθ + 2 ln |secθ + tanθ| + C recall that: secθ = x/2 hence tanθ = √(sec²θ - 1) = √[(x/2)² - 1] = √[(x²/4) - 1] = √[(x² - 4) /4] = [√(x² - 4)] /2 thus, substituting back, you get: 2tanθ secθ + 2 ln |secθ + tanθ| + C = 2 {[√(x² - 4)] /2} (x/2) + 2 ln |(x/2) + {[√(x² - 4)] /2}| + C = concluding with: ∫ [x² /√(x² - 4)] dx = (x/2)√(x² - 4) + 2 ln |[x + √(x² - 4)] /2| + C I hope it helps..
Integrate 1/x(sqrt(4x^2+9))dx using trigonometric substitution x=(3tan/2)?
INTEGRAL[1/(x*sqrt(4x^2+9)) dx] 2x=3tanu; x=(3/2)tanu; dx=(3/2)sec^2(u) du INTEGRAL[(3/2)*sec^2(u) / ((3/2)tanu * sqrt(9tan^2(u)+9)) du] INTEGRAL[(3/2)(2/3)(1/3)sec^2(u) / (tanu* sqrt(sec^2(u))) du] INTEGRAL[(1/3)cscu du] (1/3)ln(cscu - cotu)+C Now cotu=3/(2x) and cot^2(u)+1=csc^2(u) So cscu=sqrt(9/(4x^2) + 1) or sqrt(9+4x^2)/2x We get (1/3)ln[(sqrt(9+4x^2) - 3)/(2x)] + C
How to solve the integral of sqrt(9-4x^2) using trig substitution?
The directions say to use appropriate trigonometric substitution to evaluate it, but I'm not sure how to do it, as the 9 and 4 are throwing me off. Any help would be greatly appreciated!
Trigonometric Substitution ∫√(1-4x²) dx?
∫√(1 - 4x²) dx = let x = (1/2) cos u → cos u = 2x → u = arccos(2x) dx = (1/2)(-sin u) du = - (1/2)sin u du then, proceeding with substitution, ∫√(1 - 4x²) dx = ∫√{1 - 4[(1/2) cos u]²} [-(1/2)sin u] du = - (1/2) ∫√{1 - 4[(1/4) cos²u]} (sin u) du = - (1/2) ∫ [√(1 - cos²u)] (sin u) du = - (1/2) ∫ [√(sin²u)] (sin u) du = (*) - (1/2) ∫ (sin u) (sin u) du = now let (sin u) du (second factor) = dv → - cos u = v (sin u) (first factor) = U → cos u du = dU now, integrating by parts, - (1/2) ∫ (sin²u) du = - (1/2) ∫ (sin u) (sin u) du = - (1/2) [(- cos u) (sin u) - ∫ (- cos u) cos u du] = (1/2) cos u sin u - (1/2) ∫ cos²u du thus, to sum up: - (1/2) ∫ sin²u du = (1/2) cos u sin u - (1/2) ∫ cos²u du → - (1/2) ∫ sin²u du = (1/2) cos u sin u - (1/2) ∫ (1 - sin²u) du → - (1/2) ∫ sin²u du = (1/2) cos u sin u - (1/2) ∫ du + (1/2) ∫ sin²u du → now, having got the same integral at both sides, let us gather the unknown integral (1/2) ∫ sin²u du at left side: - (1/2) ∫ sin²u du - (1/2) ∫ sin²u du = (1/2) cos u sin u - (1/2) ∫ du → - (1/2) ∫ sin²u du - (1/2) ∫ sin²u du = (1/2) cos u sin u - (1/2) ∫ du → - ∫ sin²u du = (1/2) cos u sin u - (1/2) ∫ du → - ∫ sin²u du = (1/2) cos u sin u - (1/2)u + c → and therefore (see (*) above): - (1/2) ∫ sin²u du = (1/4) cos u sin u - (1/4)u + c now, substituting back u = arccos(2x), you get: ∫√(1 - 4x²) dx = (1/4) cos [arccos(2x)] sin [arccos(2x)] - (1/4)arccos(2x) + c in order to express the first term of the result in algebraic form, after letting back u = [arccos(2x)], you have to express sin u in terms of cos u, that is: sin u = √(1 - cos²u) in which cos u = cos[arccos(2x)] = 2x and therefore: sin u = sin [arccos(2x)] = √(1 - cos²u) = √[1 - (2x)²] = √ (1 - 4x²) thus your definitive result is: ∫√(1 - 4x²) dx = (1/4) cos [arccos(2x)] sin [arccos(2x)] - (1/4)arccos(2x) + c → ∫√(1 - 4x²) dx = (1/4)(2x)√(1 - 4x²) - (1/4)arccos(2x) + c = (1/2)x√(1 - 4x²) - (1/4) arccos(2x) + c I hope it has been helpful Bye!
What is the integration of square root 4-x2?
By x2, I'm going to assume you mean x^2. Integral ( sqrt(4 - x^2) dx ) You definitely have to use trigonometric substitution to solve this. I'm going to use "t" instead of "theta" though. Let x = 2sin(t). Then dx = 2cos(t) dt Applying the substitution gives us Integral ( sqrt(4 - (2sin(t))^2) 2cos(t) dt ) Simplifying, Integral ( sqrt(4 - 4sin^2(t)) 2cos(t) dt ) Factor the 4 inside the square root, and the 2 outside of the integral. 2 * Integral ( sqrt [ 4(1 - sin^2(t)) ] cos(t) dt ) The 4 comes out of the square root as 2. 2 * Integral ( 2 * sqrt(1 - sin^2(t)) cos(t) dt ) Pull the 2 out of the integral, as it is a constant. 2 * 2 * Integral ( sqrt(1 - sin^2(t)) cos(t) dt ) Multiply the 2's together. Use the trig identity 1 - sin^2(t) = cos^2(t). 4 * Integral ( sqrt( cos^2(t) ) cos(t) dt ) The square root of a squared term is itself. 4 * Integral ( cos(t) cos(t) dt ) 4 * Integral ( cos^2(t) dt ) To integrate squared trigonometric functions by themselves, we must use the half angle identity. A reminder that cos^2(t) = (1/2) (1 + cos(2t)) 4 * Integral ( (1/2) (1 + cos(2t)) dt ) Factor the (1/2) out of the integral. This merged with the 4 to become 2. 2 * Integral ( (1 + cos(2t)) dt ) And now, we can integrate term by term. A reminder that, for a constant n, the integral of cos(nt) = (1/n)sin(nt). 2 (t + (1/2) sin(2t) ) + C Distribute the 2, 2t + sin(2t) + C Apply the double angle identity sin(2t) = 2sin(t)cos(t) 2t + 2sin(t)cos(t) + C To convert this back in terms of x, we must use trigonometry and SOHCAHTOA. Since x = 2sin(t), then sin(t) = x/2 A reminder that by SOHCAHTOA, sin(t) = opp/hyp. Therefore, opp = x, hyp = 2, so by Pythagoras, adj = sqrt(2^2 - x^2) = sqrt(4 - x^2) It follows that cos(t) = adj/hyp = sqrt(4 - x^2)/2 Furthermore, since sin(t) = x/2, t = arcsin(x/2). Using all of these facts, 2t + 2sin(t)cos(t) + C becomes 2arcsin(x/2) + 2 (x/2) (sqrt(4 - x^2)/2) + C Which converts to 2arcsin(x/2) + (x/2) sqrt(4 - x^2) + C
How can I solve the integral of x^3√(9-x^2) dx using trigonometric substitution? ?
∫ x^3√(9-x^2) dx So then I know that x = 3sinθ dx = 3cosθdθ When I substitute, it becomes ∫ (3sinθ)^3 * √(9-(3sinθ)^2) * 3cosθdθ = ∫ (27sin^3θ * (3 - 3sinθ)* 3cosθdθ Is there any way to furter simplify this before I solve it? And if there isn't, how would I go about solving it?