what is the integration of (sin (ax^2 + b))dx ?
149854
babymegan101 is wrong because the derivative of xsin(ax^2+b)+C is sin(ax^2+b) + 2ax^2*cos(ax^2+b). (And we expect the derivative to be sin(ax^2 + b)). I've tried every trick in a physicist's toolbox to solve this integral but I have not succeeded. Therefore I propose that there isn't a definitive answer to the question, and that it can only be solved by approximation using numerical methods.
840169
u= x\ndv=cosx dx\n\ndu =dx\nv= sinx\n\n\nintegral of xcosx dx = xsinx - integral of sinx dx\n= xsinx + cosx +C\n`
147855
Try http://integrals.wolfram.com/
929562
∫ax(2 -x)dx\n= ∫(2ax - ax²)dx\n= 2a * ½x² - a * ⅓x³ + c\n= ax² - ⅓ax³ + c
743452
Assuming that x^4 is not in the denominator:\n\n(x-2/x^2) simplifies to  (x^-1) - (2x^-2)\n\nTherefore, the problem becomes:\n\nint[ x^-1  -  2x^-2  + x^4 ]\n\nNow you can use the power rule to integrate each term:\n\nln(x) + 2x^-1 + (x^5)/5  +  C
117158
the integration >>> increase the power by one and divide by the new power (don't forget the the constant)\n\nx^n -------> x^(n+1)/(n+1)\n\ny = 1/(x^2)\ny = x^-2\nthe integration = x^(-1)/(-1) + c\n= -x^-1 + c\n= -1/x + constant
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.
786513
put x^2=t\n2xdx=dt\nxdx=1/2dt\nthe integral now is\n(1/2dt/1+t^2\n(1/2)tan^-1t+C\n=(1/2)tan^-1(x^2)+C
932663
Let a = A cos C\nb= A sin C\n\nsquare and add a^2+b^2 = A^2\n\nthen given expression\n= A cos C sin Bx + A sin C cos Bx = A sin(Bx+C)
879050
You need to make the trig substitution, x = 2 tan(u). Then dx = 2 sec^2(u) du.\nThe integral becomes\n(1/(4tan^2(u)*sqrt(4tan^2(u)+4)))*2sec^2(u)\n\nUsing the Pythagorean identity, tan^2(u) + 1 = sec^2(u), the square root disappears and stuff cancels. If you rewrite everything in terms of sin and cos, more stuff cancels. You should get\ncos(u)/(4sin^2(u))\n\nA better way to write this is\n(1/4)(cot(u))(csc(u))\nsince this is very close to the derivative of csc(u), which is\n-cot(u)csc(u)\n\nThus the integration is -(1/4)csc(u).\nIf you draw a right triangle with an angle with measure u, the leg opposite u has length x, the leg adjacent to u has length 2, since we said above that x=2 tan(u), or equivalently, tan(u)=x/2.\nThe hypothenuse of this triangle is sqrt(x^2+4), by the Pythagorean theorem. Then csc(u) is sqrt(x^2+4)/x.\n\nThe final answer is -sqrt(x^2+4)/4x.\n\nI tried this on the Mathematica Integrator and the answer it gave matches mine. You have to input 1/(x^2*Sqrt[x^2+4]). Note that Sqrt is capitalized, to tell Mathematica that you mean Sqrt, the function, not sqrt, the variable; also note the use of square brackets (instead of rounded parentheses), which indicate the thing you're square-rooting.
482644
Let us denote the integral sign by I.\nWe will use substitution method.\nLet x = 2*tan(t)\nThen dx = 2*sec^2(t)dt\nThus, I[1/(x^2+4)^2]dx\n= I[1/{4*tan^2(t)+4}^2]*2*sec^2(t)dt\n= (1/8)*I[1/{tan^2(t)+1}^2]*sec^2(t)dt\n= (1/8)*I[1/{sec^2(t)}^2]*sec^2(t)dt  [since tan^2(t) + 1 =sec^2(t)]\n= (1/8)*I[1/{sec^2(t)}]dt\n= (1/8)*I[cos^2(t)]dt\n= (1/16)*I[2*cos^2(t)]dt\n= (1/16)*I[1+cos(2t)]dt [since 2*cos^2(t) = 1 + cos(2t)]\n= (1/16)*I[dt] + (1/16)*I[cos(2t)]dt\n= (1/16)*t + (1/16)*[{sin(2t)}/2] + C where C = arbitrary constant of integration\n= t/16 + {sin(2t)}/32 + C\n\nNow sin(2t) \n= {2*tan(t)}/{1+tan^2(t)}\n= 4x/(4+x^2) [since x = 2*tan(t)]\n\nAlso t = tan-1(x/2) [tan-1(x/2) means tan inverse (x/2)]\n\nHence the given integral equals to\n{tan-1(x/2)}/16 + x/{8(4+x^2)} + C
696867
cos[x]dx=dsin[x] therefore the integral is over e^y in the limits y=sin[Pi/2]=1 to y=Sin[0] = 0; and is equal to 1-Exp[1]
1162151
Use the fact that tan^2(x) = sec^2(x ) - 1 and integrate that.  You will get tanx -x +c.
1021739
Since tanx = sinx/cosx,\n\ntan(A-B) = sin(A-B)/cos(A-B)\n\nNow, write out the expression for each term on the right side of this equation.\n\nWhen you finish that, divide each term by cosAcosB and you will get the result you are seeking.
339966
The integral of sin(x)/x cannot be written in finite terms with the 'elementary functions', which consist of trig, inverse trig, exponential, logarithms, and algebraic functions. It can be written as a power series or in other ways, depending on what you want to do with it. Good amth programs will compute values for this integral (as approximations).\n\nAs for the integral of sin(x)/x from 0 to infinity, the result is pi/2, but that usually is done with complex analysis. It cannot be done easily with elementary methods.
859250
integral = x*tan(x)+ln(sec(x))  (use integration by parts with du = sec^2(x) and v = x).\n\ng'(x) = f(x), so a, c, e and g are true (b is true if f is continuous)\n\nSteve
744903
1)\n\ntan^2(x) = sec^2(x) - 1\n\nSo the integral of sec^2(x) + 4 = tan(x) + 4x\n\n2)\n\ncsc(x) = 1/sin(x)\n\nthe integral is then of 1+sin(x)/cos^2(x).\nlet u = cos(x) then du = -sinx dx.\nso the integral is x+sec(x).\n\n3)\n\nlet u = cos(2x+1) then du = -(1/2)sin(2x+1) dx.\n\nso the integral is of -(1/2)/u^2 = 2/u = (1/2)/cos(2x+1) = sec(2x+1)/2\n\nSteve