Y=0..1, x = 0..3 integral of 4xy sqrt (x^2 + y^2) dy dx?
How do you solve this double integral? Here is the answer: integral_0^3 integral_0^1 4 x y sqrt(x^2+y^2) dy dx = 16/15 (25 sqrt(10)-61)~~19.2607 Can you please show every step clearly. Don't skip on things that should be assumed to be known, because I feel like I'm making a small mistake that I can't catch. Thanks
What is the Integral of x^3 * sqrt(1-x^2) dx?
Hey y'all, I am trying to solve this problem and it it lists as being needed to be done with forms of trig identities but I am getting a little stuck. I ended up setting x as sin(u) , made it forward to the integral of sin^3 u * cos^2 u du, and then to the integral of sin^3 u - sin ^5 u du but now it seems I am stuck! Help please!
Integral: x/(sqrt(x^2+2x+10)) dx?
Use trigonometric substitution which is explained with a link in the source. ∫ x / √(x² + 2x + 10) dx ∫ x / √[(x + 1)² + 9] dx Let x = 3tanθ - 1, 3tanθ = x + 1 tanθ = (x + 1) / 3 θ = tan‾¹[(x + 1) / 3] dx / dθ = 3sec²θ dx = 3sec²θ dθ ∫ (3tanθ - 1) / √(9tan²θ + 9) 3sec²θ dθ ∫ (3tanθ - 1) / √[9(tan²θ + 1)] 3sec²θ dθ ∫ (3tanθ - 1) / √(9sec²θ) 3sec²θ dθ ∫ (3tanθ - 1) / 3secθ 3sec²θ dθ ∫ secθ(3tanθ - 1) dθ ∫ (3secθtanθ - secθ) dθ 3 ∫ secθtanθ dθ - ∫ secθ dθ 3secθ - ln|secθ + tanθ| + C Since θ = tan‾¹[(x + 1) / 3], 3sec[tan‾¹((x + 1) / 3)] - ln|sec[tan‾¹((x + 1) / 3)] + tan[tan‾¹((x + 1) / 3)]| + C 3√[(x + 1)² / 9 + 1] - ln|√[(x + 1)² / 9 + 1] + [x + 1] / 3| + C 3√[((x + 1)² + 9) / 9] - ln|√[((x + 1)² + 9) / 9] + [x + 1] / 3| + C √[(x + 1)² + 9] - ln|√[(x + 1)² + 9] / 3 + [x + 1] / 3| + C √[(x + 1)² + 9] - ln|[√[(x + 1)² + 9 + x + 1] / 3| + C √[(x + 1)² + 9] - ln|√[(x + 1)² + 9] + x + 1| - ln3 + C √[(x + 1)² + 9] - ln|√[(x + 1)² + 9] + x + 1| + C √[(x + 1)² + 9] - ln|x + √[(x + 1)² + 9] + 1| + C √(x² + 2x + 10) - ln[x + √(x² + 2x + 10) + 1] + C
How do I integrate the square root of (1-x^2)/x?
Assuming it is indefinite integrationTake x=sinθIf you solve the equation it becomes sqrt(cosθ)^2 / sinθ which is equal to cotθIntegral of cotθ is equal to ln(sinθ) + CRefer Table of Integrals for integration value
Find the integral ∫(sqrt x)/(sqrt 1+x^3) dx?
Start out by substituting: tan(u) = x^(3/2) ==> x = [tan(u)]^(2/3) and dx = (2/3)sec^2(u)[tan(u)]^(-1/3) du. By applying these substitutions ∫ √x/√(1 + x^3) dx = 2/3 ∫ sec^2(u)[tan(u)]^(-1/3)[tan(u)]^(1/3)/ √[1 + tan^2(u)] du = 2/3 ∫ sec^2(u)[tan(u)]^(-1/3)[tan(u)]^(1/3)/ sec(u), since sec^2(u) = 1 + tan^2(u) = 2/3 ∫ sec(u) du, as all other terms cancel = (2/3)ln|sec(u) + tan(u)| + C, by integrating = (2/3)ln|sec[x^(3/2)] + tan[x^(3/2)]| + C, by back-substituting u = x^(3/2). I hope this helps!
Indefinite integral of x^3 sqrt(x^2 + 1) dx?
Your step by step answer is here : http://s636.photobucket.com/albums/uu88/... Original link : http://www.wolframalpha.com/input/?i=x^3... *******