Does anybody know the integral of (Cscx)^3 dx ?
468225
Write (cscx)^3 as (cscx)(cscx)^2\n(cscx)^2 = 1 + (tanx)^2\n\n(cscx)(cscx)^2\n= (cscx)(1 + (tanx)^2)\nDistribute cscx\ncscx + (cscx)(tanx)^2\n\nSimplify (cscx)(tanx)^2\n(cscx)(tanx)^2 = (1/sinx)((sinx)^2/(cosx)^2)\n= (sinx)/(cosx)^2\n= (1/cosx)(sinx/cosx)\n= (secx)(tanx)\n\nThus, \ncscx + (cscx)(tanx)^2\n= cscx + (secx)(tanx)\nThe antiderivative of cscx is -ln|cscx + cotx| \nThe antiderivative of (secx)(tanx) is secx.\nHence, the integral of (Cscx)^3 is\n-ln|cscx + cotx| + secx + C, where C is a constant.
502365
Not as stated.\nIt should be;\n\nThe integral of z-squared d z\nbetween one and the cube root of three\ntimes the cosine\nof 3 pi over nine\nis the log of the cube root of e\n\n\nThe integral of z-squared d z\nbetween one and the cube root of three = 2/3\n\ncos(3pi/9) = cos(pi/3) = 1/2\n\nlog (natural) cube root of e = 1/3
414104
Using an integral,\n\nWords = S 3t(3t^2 + 1)^(-1/2) dt\n\nlet u = 3t^2 + 1\nthen du = 6t dt\n\n= (1/2) S u^(-1/2) du\n= (1/2)(2u^(1/2))\n= u^(1/2)\n= (3t^2 + 1)^(1/2)\n\nEvaluating from 1 to 4,\n= (3(4)^2 + 1)^(1/2) - (3(1)^2 + 1)^(1/2)\n= 49^(1/2) - 4^(1/2)\n= 7 - 2\n= 5\n\nAnswer A.
700700
try putting x^2 = i*t\nwhere i = complex iota\n t is varible \n\nthere integral will be in form : i/2*( cos(t)+i*sin(t) ) dt\n\nso answer will be : 1/2 * (i* sin (t) + i*cos(t) ) = 1/2 * e^ (i*t) = 1/2 * e^x^2\n\nThat's the answer....... :D\n\n\n[Assuming u know that e^(i*t) = cos(t)+i*sin(t) ]\n\ncan be solved easily ......... try it
1068054
well Y' = dy/dx\nso using seperation of veriables we can get\ndy/dx = K - 1/3*y^3\nand dy/(K-1/3*y^3)=dx now integrate both sides. \nDoes that help?
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
200165
I will assume cos(x)^3 means (cos x)^3 and does NOT mean  cos(x^3)\nFirst start by separating the integral into two pieces.\nI will use "int" to stand for the integral sign.\nint{(cosx)^3/(cosx)^2}dx - int{(sinx)^2/(cosx)^2}dx\n\nNow (cosx)^3/(cosx)^2 = cosx\nand (sinx)^2/(cosx)^2 = (tanx)^2 = (secx)^2 - 1\nThus, the integral is now equal to \n\nint{cosx}dx - int{(secx)^2 - 1}dx\n\nThe antiderivative of cosx is sinx.\nThe antiderivative of (secx)^2 is tanx.\nThe antiderivative of -1 is -x.\n\nHence, the solution is \nsinx - tanx + x + C\n\n(The signs changed for tanx and -x because there was a negative sign in front of their integral, and the negative sign has to be distributed).
108152
(1/3)*e^(3x)-4sinx + c, where c is any arbitrary constant.
900298
We know that F(x) = ∫ln³ x/x dx + C. So we need to find that integral, solve for the constant of integration, and then plug 9 into the resulting formula. First, the integral:\n\n∫ln³ x/x dx\nLet u=ln³ x, v=ln x, du=3 ln² x/x dx, dv = 1/x dx. Integrating by parts:\n∫ln³/x dx = ln^4 x - ∫3 ln³ x/x dx + C\nAdding 3∫ln³ x/x dx to both sides:\n4∫ln³/x dx = ln^4 x + C\nDividing by 4:\n∫ln³/x dx = ln^4 x + C\n\nNow, solving for C:\n\nF(1) = ln^4 1 + C\nF(1) = 0^4 + C\nF(1) = C\nBut F(1) = 0, so C=0.\n\nNow, solving for F(9):\n\nF(9) = ln^4 9 = (ln 9)^4 ≈ 23.30761
772048
cot²(x)\n------------ =  ............ (NOTE cot²x + 1 = csc²x)\ncsc(x) - 1\n\ncsc²(x) -1\n------------ =\ncsc(x) - 1\n\n(csc(x) - 1)(csc(x) + 1)\n----------------------------- =\ncsc(x) - 1\n\ncsc(x) + 1
748051
The primitive of your integral is:\n\ns*(e^(st))*(1/w)*atan(s/w)\n\nYou have to do a trigonometric substitution to get there. Do not always trust the weird answers you get with complex calculators.
761613
Using implicit differentiation, you get:\n\n-csc^2(y)(dy/dx) = 1 - dy/dx\n-csc^2(y)(dy/dx) + dy/dx = 1\n(dy/dx)[1 - csc^2(y)] = 1\n\nTherefore, dy/dx = 1/[1 - csc^2(y)]\n\nNow plug in your ordered pair to get dy/dx.  Remember csc (y) = 1/sin(y):\n\ndy/dx = 1/[1-1], which is undefined.  Therefore, you cannot write a tangent line for this ordered pair (there is a vertical asymptote at this point)\n\n------------\n\nHope this helps
480262
the integration for any intgral from (a) to (-a) is always equal zero.\nso the intgral for this is zero\nbut if there is no interval the integration is very complicated
527554
Let I=Int((e^5x)*cos 3x)dx               (1)\n\nUse the formula for integration by parts twice\n\nInt(u dv/dx)dx= uv - int(v du/dx)dx\n\ntaking u = e^5x   and  dv/dx=cos 3x\n   du/dx=5(e^5x)  and      v=(1/3)*sin 3x\n\nthen I = (1/3)*e^5x(sin 3x) - (5/3)*int((e^5x)*sin 3x)dx      (2)\n\nUse integration by parts again: -\n\nint((e^5x)*sin 3x)dx = (-1/3)*(e^5x)*(cos 3x)+(5/3)*int((e^5x)*sin 3x)dx\n\nNow the int((e^5x)*sin 3x)dx we have already defined to equal I\n\ntherefore\n\nint((e^5x)*sin 3x)dx = (-1/3)*(e^5x)*(cos 3x)-(5/3)*I\n\nputting this into the equation 2 to get: -\n\nI = (1/3)*e^5x(sin 3x) + (5/9)*(e^5x)*(cos 3x)-(25/9)*I\n\nBring I over to the left hand side: -\n\n(34/9)I = (1/3)*e^5x(sin 3x) + (5/9)*(e^5x)*(cos 3x)\n\nWith a bit of careful rearranging, you can get the answer :-\n\n    I = (1/34)*(e^5x)(3sin3x+5cos3x) + C\n\nwhere C is an arbitrary constant\n\nI know it looks a bit of a bee-hatch written out here, but go through it and make sure you understand what has been done at each step. I may have made a few errors involving minus signs, so make sure you understood what i have done.\n\nI got the answer wrong first time round, but i have gone through it again and corrected the mistakes.
734906
the integral of [cos 2x dx] is [1/2 sin 2x + c]
794871
assuming 0 and 3 are limits of integration:\nv(t) = dx/dt = t^2 - 2t\nx = (1/3)t^3 - t^2 + C\nx = (1/3)3^3 - 3^2 - 0 + 0\nx = 9 - 9 = 0\n\nIf the limits are 0 and 2, \nx = 8/3 - 4 = -4/3\n\nGoing backwards,\n(1/3)t^3 - t^2 = 8/3\nt^3 - 3t^2 -8 = 0\nt ≈ 3.612887865
706536
Integral: ∫ [2/{(e^2x)+4}] dx\n\nFirst consider the expression: 2/{(e^2x) + 4}\n\n= {e^(2x) + 4}/[2*{e^(2x) + 4}] - e^(2x)/[2*{e^(2x) + 4}]\n\n= 1/2 - (1/2)[e^(2x)/{e^(2x) + 4}]\n\n\nHence, ∫[2/{(e^2x) + 4}] dx\n\n=  ∫[(1/2) - (1/2)[e^(2x)/{e^(2x) + 4}]] dx\n\n=  ∫(1/2) dx - ∫(1/2)[e^(2x)/{e^(2x) + 4}] dx\n\n=  (1/2) x - (1/2)∫[e^(2x)/{e^(2x) + 4}] dx\n\nLet {e^(2x) + 4} = z. Then dz = 2e^(2x) dx\n\nHence the given integral becomes:\n(1/2) x - (1/4)∫(dz/z)\n\n= (1/2) x - (1/4)ln|z| + c where c = arbitrary constant of integration\n\nNow, substitute the value of 'z' in terms of 'x', thus the given integral becomes\n= (1/2) x - (1/4)ln|e^(2x) + 4| + c