how to integrate sin(x)^2/x^2 from zero to inf?
913797
You can begin by converting sin x^2 to 1-cos2x and since they have the same denominator x^2 just split it in two integrals (1/x^2) - (cos2x/x^2).\nThe first one is easy to solve, do initegration by parts for the second integral.
836562
f'(x)=2-5sinx\nsetting this to zero\nsinx=2/5\nabsolute extreme=2(2/5)+cosarcsin(2/5)
727341
first integrate\n\nF(x) = x^3 -x^2  from [1, 3]\n= 3^3 -3^2 -1^3 + 1^2\n= 27 - 9 -1 + 1\n= 18\n\naverage = 18/(3-1) \n= 18/2\n= 9
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.
294550
Graph sin(2x) and sin(x). Once graphed you will see that y=sin(2x)the top function. You will integrate (sin2x-sinx)dx from 0 to pi. you can split the intergrals\nint(sin2x)dx = -1/2cos2x \nminus\nint(sinx)dx = -cosx\n\n-1/2cos2x+cosx evaluated at pi and 0.\n\n(-1/2 cos(2(pi))+ cos(Pi))-(-1/2 cos(0)+cos(0))
730722
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck
120506
i'm giving the final answer not hte full steps ok\n(a sin(wt)-w cos(wt))/w*(w^2+a^2)  + e^(-at)/(a^2+w^2)\nby using convolution theorem
734906
the integral of [cos 2x dx] is [1/2 sin 2x + c]
274355
Depends on the system you are trying to solve. check your limits of integration. from -infinity to 0 plus 0 to infinity.
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
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
904086
I would use substitution method on this one...\n\nFirst, break it up into the sum of 2 integrals...\n\nInt(cost/(1+sin^2t))dt + Int(-2sint/(1+cos^2t))dt\n\nFor the first integral let u = sint\n\ndu = cost\n\nThen you have an integral that you can evaluate ...\n\nInt(1/(1 + u^2)du = arctan(u)\n\nSubstitute cos(t) back in for u and you get arctan(cos(t)), evaluate that from 0 to pi/2.\n\nFollow the same reasoning for the 2nd one!
690460
Integrate sqrt(r^2-x^2) from0 to r-h and 0 to r. (Integrate 2 times )
1183491
Lim { [ sin(x-1)  ] / [ x^2 + x - 2] }\nx-->1\n\nAdd the following to the denominator\n-2x + 2x +1 -1\n\nLim { [ sin(x-1)  ] / [ x^2 + x - 2 -2x +2x +1 -1] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ x^2 -2x + 1 +3x -3] }\nx-->1\n\nNote: (x^2-2x+1) =(x-1)^2    and  (3x-3) = 3(x-1)\n\nThen\nLim { [ sin(x-1)  ] / [ (x-1)^2 +3(x-1)] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ (x-1)^2 +3(x-1)] }\nx-->1\n\nLim { [ sin(x-1)  ] / [ (x-1)[ (x-1)+3 ] }\nx-->1\n\nLim { [ sin(x-1) / (x-1) ] * [ 1/ (x-1+3 )] }\nx-->1\n\nLet u= x-1 then as x-->1, u--> 0\nand\nLim { [ sin u / u ] * [ 1/ (u+3 )] }\nu-->0\n\nSince\n\nLim  [ sin(u) / (u) ]=1\nu-->0\n\nThen \nLim { [ sin u / u ] * [ 1/ (u+3 )] } = 1 * [1/(0+3)] = 1*1/3 = 1/3\nu-->0
704235
I would expand (sin 2x+1)^2, then replace (sin 2x)^2 by\n(1-cos 4x)/2, then integrate term by term.  If you can't do that, I expect another answerer will do the whole thing, or else you can email h_chalker@yahoo.com.au
965420
You have to use trigonmetric substitution, which is by far, the hardest type of integration.  It also involves making diagrams of triangles, which I hope you will bear with me with this limited interface.\n\nSo we want to solve for Integral (1/(1+y^2))dy.  You have to use trigonometric substitution, and \n\nlet y = tan(x).  Then dy = (sec(x))^2, or "secant squared x", and we just directly substitute that.\n\nIntegral (1/(1+y^2))dy = \nIntegral (1/(1 + (tan(x))^2) (sec(x))^2 dx =\n\nNote:  In trigonometric substitution, unlike the regular type, you're also replacing dy for what you solved for.\n\nUsing the trig identity (sec(x))^2 = (tan(x))^2 + 1, we get\n\nIntegral (1/(sec(x))^2 * (sec(x))^2) dx = (the two values will cancel each other out into 1)\nIntegral (1)dx =\n\nx + C\n\nRemember that above, we let y = tan(x), so we have to solve for x, and we get x = tan(inverse)y, or tan(x) with a -1 value between the "tan" and the "x".  The reason for this step is we require an answer __in terms of y__.\n\nSo, our actual answer is tan(inverse)(y) + C
808989
Let t=sinx\n=> dt = cosx dx\n\nx=pi/2 => t=1\nx=0 => t=0\n\nI=integrate(e^t dt) from 0 to 1\n\n=e^t from 0 to 1\n\n=> I= e^1 - e^0 = e-1\n\nAnswer is I=e-1