pls help me in solving for : integral Arcsin(x) / x^2 dx?
820456
I believe Integration by Parts would be your best bet here, since (a) it's straightforward to differentiate Arcsin(x), and (b) it's not hard to integrate x^(-2).\n\nLet u = Arcsin(x) and let dv = x^(-2) dx\nThus du = 1/sqrt(1-x^2) dx and v = -1/x\n\nSet up the IBP formula:\n\n Int [Arcsin(x)/x^2 dx] = uv - Int [v du]\n  =  Arcsin(x)/x - Int [-1/(x sqrt(1-x^2)) dx]\n  = Arcsin(x)/x + Int [ x^(-1) (1-x^2)^(-1/2) dx]\n... (by substitution*)\n  =  Arcsin(x)/x + ln(x) - ln (sqrt (1-x^2) + 1)\n\nBe SURE to visit the Integrator at Wolfram.com (see link below)!\n\n* See also worked-out example #6 at:\nhttp://mathforum.org/mathtools/cell/c,15.14.6,ALL,ALL/
756591
Given\n\nsin 3x = 1/2\n\n\n\nGet arcsin of both sides\n\narcsin (sin 3x) = arcsin (1/2)\n\n\n\nThe left side simplifies as 3x.  The right side varies.  \nLet k be any integer.  Thus,\n\n3x = π/4 + 2kπ and\n3x = 3π/4 + 2kπ\n\n\n\nThus,\n\n3x = (π + 8kπ)/4 and\n3x = (3π + 8kπ)/4\n\n\n\nTherefore,\n\nx = π/12 (8k + 1) and\nx = π/12 (8k + 3)\nwhere k is ANY integer. Enumerating, we have,\n\nx = ... -5π/4, -7π/12, π/12, 3π/4, 17π/12, 25π/12, 11π/4, ...\nx = ... -13π/12, -5π/12, π/4, 11π/12, 19π/12, 9π/4, 35π/12, ...\n\n^_^
691116
x/[a^(-x^2/(2a))] dx\nRewrite bringing the denominator to the top.\nThis will change the exponent to positive.\nx(a^(x^2/(2a)) dx\nLet u = x^2/(2a), then du = (1/a)x dx or \na du = x dx\nNow the integral is\nint(a(a^u) du)\n= int(a^(u+1) du)\n= a^(u+1) + C\n= a^(x^2/(2a) + 1) + C\n(The exponent of a is x^2/(2a) + 1)
697906
let y = arc tan (x)\n\ntan y = x\ndifferentiate\n\nsec^2 y dy/dx = 1\n\nor(1+tan ^2y) dy/dx = 1\n\ndy/dx = 1/(1+tan^2 y) = 1/(1+x^2)
889651
It's just arctan (x^2).\n\nBy the fundamental theorem of calculus, if F(t) is the antiderivative of f(t) = arctan(t^2), then the integral of f(t) from 0.5 to x will give F(x) - F(0.5). The derivative of this function with respect to x then gives you f(x).
574666
int((x-2)*(x-5)^(1/2)*dx)\n\nUse u sub, let:\nu=x-5\ndu=dx\nx-2=u+3\n\nResub:\n\nint((u+3)*sqrt(u)*du)\n\nDistribute:\n\nint((u^3/2 + 3sqrt(u)*du))\n\nIntegrate:\n\n0.4*u^(5/2) + 2*u^(3/2) + c\n\nResub u for x:\n\n0.4*(x-5)^(5/2) + 2*(x-5)^(3/2) + C
825804
First, let's simplify sin(2x) / sin(x)\nRemember that sin(2x) = 2sin(x)cos(x), so\nsin(2x) / sin(x) \n= 2sin(x)cos(x) / sin(x)\n= 2cos(x)\nThe integral of 2cos(x) dx\n= 2sin(x) + C
855519
I'm sure someone else will do better than I will on this, but if it helps, according to the table of integrals in my calc textbook, it should be:\n\nx²/2 √(a² - x²) + a²/2 sin-¹(x/a) + C\n\nThat's sin^-1, aka arcsin.\n\nI'm way past the point where I could explain how to derive it without reviewing unfortunately.
935281
The general formula is\narctan(x)-arctan(y)=\narctan[(x-y)/(1+xy)].\nIn your case, x=1+sqrt(2), y=1-sqrt(2), so\nx-y=2sqrt(2) and xy=-1. Thus, you get \narctan(2sqrt(2)/0)=pi/2.
882948
Just make the following substitution:\nr=sqrt(x^2+y^2)\ntheta=arctan(y/x)\nand then simplify.  Bring everything over to one side and you get your answer.\n\nIf tan(theta)=y/x, then sec(theta)=sqrt(x^2+y^2)/x.\n\nThe problem is a little ugly, but it is doable and then at the end everything ends up canceling nicely.
829669
integral=4x^-3dx-x^-2dx\n=4(x^-2)/-2-x^-1/-1\n=-2/x^2+1/x+C
1054686
Job one, convert everything to trig functions of a single angle, namely x.\n\nsin2x=2sinxcosx\ncos2x=cos^2(x)-sin^2(x)=2cos^2(x)-1=1-2sin^2(x)\n\nThen substitute in and simplify:\n2sinxcosx+2(2cos^2(x)-1)=1+sinx-4cosx\n2sinxcosx+4cos^2(x)-2-1-sinx+4cosx=0\n4cos^2(x)+2sinxcosx-sinx+4cosx-3=0\n\nFactor this into:\n(2cosx-1)(2cosx+sinx+3)=0\n\nset each parantheses equal to zero\n2cosx-1=0   \ncosx=1/2\nx=arccos(1/2)\nx={pi/3, 5pi/3} + 2pi*n  where n=0,1,2....\n\nAnd\n\n2cosx+sinx+3=0\n\nSolved numerically gives imaginary solutions of\n{-i*ln((-2/5)-(i/5),-i*ln(-2-i)}\n\nSolved....
1156022
Start by letting I = integral of [sin(x)*sin(nx)]\nUse integration by parts twice.\nLet u = sin(x) and dv = sin(nx)\nAfter using integration by parts twice, you'll end up returning to the original integral you started with, so you will have to solve for I.\n(Your instructor should have shown you this trick)\n\nAnswer: [cos(x)*sin(nx) - n*sin(x)*cos(nx)]/[n^2 - 1]\nI did this quickly, so you should check my answer by working it out. \nIf you get stuck, send me an e-mail or add additional details.
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
391478
put x^2 + 2x + 2 = v\n\nThen dv/dx = 2x+2 \n\n(2x-5)/x^2 + 2x + 2) = (2x+2)/ (x^2 + 2x + 2) - 7/(x^2+2x+2)\n\n1)Integral (2x+2)/x^2+2x+2)dx= Int ([dv/dx)/v]dx =\n\nInt(dv/v) = Ln v = ln(x^2 + 2x + 2)\n\n2)Int[7/(x^2 + 2x + 2)dx] =?\n\nPut x+1 = t => dx = dt\n\nAlso, x^2 + 2x + 2 = (x+1)^2 + 1 = t^2 +1\n\nSo, Int[ 7/(x^2+2x+2)]dx = Int [7dt/(t^2+1] =\n\n= 7 Int [dt/(t^2+1)] = 7 arctan t = 7 arctan (x+1)\n\nInt [ f(x)dx] = 1) - 2) = ln(x^2 + 2x + 2)- 7arctan (x+1)\n\nFinal answer:\n\nln(x^2+2x+2) - 7arctan(x+1) + C\n\nHope this helped.
839872
Okay lets try this one ;\n\n2sin²  – 3sinx  + 1 →\n( sin x – 1)(2sin x – 1) = 0\nSin x – 1 = 0  →  sinx = 1 → x =∏ /2 \nX = 2∏ +∏ /2 , …., x = 2k∏ +∏/2 , ( k Є Z ) \n2sin x – 1 = 0  → sinx = 1/2 → x = ∏/6\nAs sin (∏ -∏/6) = sin ∏/6 = 1/2 \nSo ; \nX = 2∏ + ∏/6 , x = 4∏ + ∏/6 , …., x = 2k∏ + ∏/6 , ( k Є Z) \nX = 2∏ + 5∏/6 , x = 4∏ + 5∏/6 ,….., x = 2k∏ + 5∏/6 , ( k Є Z)\n\nGood Luck dear
887482
y=arccos(x - 5) \nDomain is 0 ≤ (x - 5) ≤ π      ie 5 ≤ x ≤ π + 5\nRange -1 ≤ y ≤ 1\n\nSo inverse function is\n\nx = arccos(y - 5)\n\nRange is 5 ≤ y ≤ π + 5\nDomain is -1 ≤ x ≤ 1\n\nie y - 5 = cosx \nSo y = cosx + 5\n\nDomain is -1 ≤ x ≤ 1\nRange is 5 ≤ y ≤ π + 5