Integral 1/sqrt(4x-x^2)?
integral 1/sqrt(4 x-x^2) dx For the integrand 1/sqrt(4 x-x^2), complete the square: = integral 1/sqrt(4-(x-2)^2) dx For the integrand 1/sqrt(4-(x-2)^2), substitute u = x-2 and du = dx: = integral 1/sqrt(4-u^2) du The integral of 1/sqrt(4-u^2) is sin^(-1)(u/2): = sin^(-1)(u/2)+constant Substitute back for u = x-2: = sin^(-1)((x-2)/2)+constant Which is equivalent for restricted x values to: = (2 sqrt(x-4) sqrt(x) log(2 (sqrt(x-4)+sqrt(x))))/sqrt(-(x-4) x)+constant
How do you integrate sqrt(6x - x^2)?
∫ √(6x - x²) dx = first of all, you have to complete the square by adding and subtracting 9: ∫ √(9 - 9 + 6x - x²) dx = ∫ √[9 - (x² - 6x + 9)] dx = ∫ √[9 - (x - 3)²] dx = substitute (x - 3) = 3sinθ sinθ = (x - 3)/3 θ = arcsin[(x - 3)/3] d(x - 3) = d(3sinθ) → dx = 3cosθ dθ yielding: ∫ √[9 - (x - 3)²] dx = ∫ √[9 - (3sinθ)²] 3cosθ dθ = ∫ √(9 - 9sin²θ) 3cosθ dθ = factor out 9: ∫ √[9(1 - sin²θ)] 3cosθ dθ = ∫ 3√(1 - sin²θ) 3cosθ dθ = replace (1 - sin²θ) with cos²θ: ∫ 3√(cos²θ) 3cosθ dθ = ∫ 3cosθ 3cosθ dθ = 9 ∫ cos²θ dθ = recall the power-reducing identity: cos²θ = [1 + cos(2θ)] /2 yielding: 9 ∫ {[1 + cos(2θ)] /2} dθ = break it up factoring the constant out: (9/2) ∫ dθ + (9/2) ∫ cos(2θ) dθ = (9/2)θ + (9/2) (1/2)sin(2θ) + C = recalling double-angle identities, (9/2)θ + (9/2) (1/2)(2sinθ cosθ) + C = (9/2)θ + (9/2)sinθ cosθ + C now recall that: θ = arcsin[(x - 3)/3] sinθ = (x - 3)/3 hence: cosθ = √(1 - sin²θ) = √{1 - [(x - 3)/3]²} = √{1 - [(x - 3)²/9]} = √{[9 - (x² - 6x + 9)]/9} = √[(9 - x² + 6x - 9)/9] = [√(6x - x²)] /3 thus, substituting back, you have: (9/2)θ + (9/2)sinθ cosθ + C = (9/2)arcsin[(x - 3)/3] + (9/2) [(x - 3)/3] {[√(6x - x²)] /3} + C = (9/2)arcsin[(x - 3)/3] + (9/2)[(x - 3)/9]√(6x - x²) + C = concluding with: ∫ √(6x - x²) dx = (9/2)arcsin[(x - 3)/3] + (1/2)(x - 3)√(6x - x²) + C I hope it helps..
Integral dx/sqrt(8x-x^2)?
The only way I can tell you your mistake is if you show me the steps that you took to get to your answer. Anyways, here's how to evaluate the integral just for your own reference: ∫dx/√(8x - x²) ∫dx/√-(x² - 8x) ∫dx/√-[(x - 4)² - 16] ∫dx/√(16 - (x - 4)²) u = x - 4 du = dx ∫du/√(16 - u²) Substitute u = 4*sin(t) ∫4*cos(t) dt/4*cos(t) = t + C = arcsin(u/4) + C =arcsin([x - 4]/4) + C
What is the Integral of (sqrt(x^2-1))/(x^4)?
integral sqrt{x^2 - 1} / x^4 dx the problem is the two-term quantity locked inside a square root, eliminate it by forcing a trig substitution and invoking an identity that collapses it to be one term. From experience, we understand that x = sec(u) works dx = sec(u) tan(u) du , compute this derivative yourself to verify this is so. then for the integrand term: x^2 - 1 = sec^2(u) - 1, but sec^2(u) - 1 = tan^2 ( u) by trig identity, so sqrt{x^2 - 1} = sqrt{tan^2(u)} = tan(u) also recall x = sec(u) --> x^4 = sec^4 (u) in the denominator of the integrand and we said l dx = sec(u) tan(u) du replace it all in the integral: sqrt{x^2 - 1} / x^4 dx = tan(u) / (sec^4) ( sec(u) tan(u) du) = tan^2(u) / sec^3(u) du = (sin^2(u) / cos^2(u) ) (cos^3(u)) du, that is, I am replacing tan(u) and sec(u) by their equivalent forms = sin^2(u) cos(u) du so the integral we have to solve is integral sin^2(u) cos(u) du integrate by parts (write this out if you cannot do it in your head or otherwise follow what I am doing) integral sin^2(u) cos(u) = sin^2(u) sin(u) - integral sin(u) 2 sin(u) cos(u) du integral sin^2(u) cos(u) = sin^3(u) - 2 integral sin^2(u) cos(u) 3 integral sin^2(u) cos(u) = sin^3(u) integral sin^2(u) cos(u) = (1/3) sin^3(u) this is the answer but we want it to be in terms of x, recall x = sec(u) = 1 / cos(u), i.e. 1 / x = cos(u). This implies cos(u) = adjacent / hypotenuse = 1 / x --> adjacent = 1, hypotenuse = x then the third leg of the "triangle" subtended by an "angle" u is opposite = sqrt{x^2 - 1} by pythagorean theorem. Thus, sin(u) = opposite / hypotenuse = sqrt{x^2 - 1} / x so sin^3(u) = (x^2 - 1)^(3/2) / x^3 Recall our result just above integral sin^2(u) cos(u) = (1/3) sin^3(u) noting that the left-hand side (see above when we first started substituting) is the original integral, i.e. integral sin^2(u) cos(u)= integral sqrt{x^2 - 1} / x^4 dx, we have integral sin^2(u) cos(u) = integral sqrt{x^2 - 1} / x^4 dx = (1/3) sin^3(u) integral sqrt{x^2 - 1} / x^4 dx = (1/3) (x^2 - 1)^(3/2) / x^3.........[Ans.]
Integrate Sqrt[(36*x^2) + (36*x^4)]?
∫ √ (36x² + 36x⁴) dx = you have first to factor out 36x² as: ∫ √ [36x² (1+ x²)] dx = ∫ 6x √(1+ x²) dx = now let (1+ x²) = u → d(1+ x²) = du → 2x dx = du then substitute: ∫ 6x √(1+ x²) dx = ∫ 3(2x) √(1+ x²) dx = 3 ∫ √(1+ x²) 2x dx = 3 ∫ √u du = 3 ∫ u^(½) du = 3 [u^(½⁺¹)] /[(1/2)+1] + c = 3 [u^(3/2)] /(3/2) + c = 3(2/3) √u³ + c = 2u√u + c finally, substituting back u = (1+ x²), you get: ∫ √ (36x² + 36x⁴) dx = 2(1+ x²)√(1+ x²) + c now, having got the antiderivative, evaluate the definite integral from 0 to1, that is: [2(1+ 1²)√(1+ 1²)] - [2(1+ 0²)√(1+ 0²)] = 2(2)√2 - 2(1)√1] = 2(2√2 - 1) Bye!
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... *******