Evaluate this integral....?
821031
I=I1+I2; I1=int{for x=pi/6 until pi/2} of x*dx = x^2/2{for x=pi/6 until pi/2}= pi^2/9;\nI2=int{for x=pi/6 until pi/2}of  25*dx/(sinx)^2; \nLet t=tg(x), then 1/(sinx)^2=1/t^2+1 and dx=(1+t^2)*dt;\nI2=int{for t=1/sqrt(3) until +inf}of 25*t^(-2)*dt = -25/t = 25*sqrt(3)
860131
It's undefined at the endpoint of the interval (pi/2).\nFirst go ahead and integrate it like you normally would.\nintegral of sin(x)/sqrt(cosx) dx\nUse substitution.  Let u = cos x, then du = - sinx dx\n= integral of -du/ sqrt(u) \n= integral -u^(-1/2) du\n= -u^(-1/2 +1) / (-1/2 + 1) \n= -u^(1/2) / (1/2)\n= -2u^(1/2)\n= -2 sqrt(u)\nReplace u with cosx\n= -2 sqrt(cos x)\nNow plug in pi/2\n- 2 sqrt(cos pi/2)\n= -sqrt(0)\n= 0\nPlug in pi/4\n-2 sqrt(cos pi/4)\n= -2sqrt( sqrt(2) / 2)\nNow subtract them\n0 - (-2sqrt(sqrt(2) /2))\n= 2sqrt( sqrt(2) / 2)
759209
This is a non-elementary integral in x. However, from the way you phrased the problem, I would guess that you want this to compute a double integral: first with repspect to x, then with respect to y. The best way of doing *that* would be to switch the order of integration. Then the integral with respect to y will be xe^(y/x). Be sure to change the limits of integration so that you are integrating over the same region in the xy-plane!
770990
I don't understand your statement of the problem.  It sounds like it might be an exercise in integral calculus, in which you integrate the expression at hand, evaluate the resulting function at the two limits, and take the difference.
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
989233
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.
32210
integral of sec(x) = ln(|sec(x) + tan(x)|)\n\n\nsec(x) = (sec(x)(tan(x) + sec(x))/(tan(x) + sec(x))\n\n= (sec(x)tan(x) + sec(x)^2)/(tan(x) + sec(x))\n\nif u = (tan(x) + sec(x)), then du = (sec(x)tan(x) + sec(x)^2) \n\nso the integral is the integral of du/u = ln(|u|)
297440
Are you looking for indefinite  integral   of x^(sin(1/x)) ?\n\nMathematica has a web site you can plug-in the integrant and get the result (if it exist in closed form)\n\nBy hand: try to break this power expression using log and exp functions. If  a=ln(b), then b=exp(a) and therefore b=exp(ln(b)).\n\nIn this way your integrant could be written as \n\nexp(ln(integrand))\nand since it is of power form, use the properties of ln and exp functions to break it down and thus reduce it to simpler integrals.
473554
You have to use integration by parts.\nLet u = ln(x) and dv = dx, then\ndu = (1/x)dx and v = x\nint(udv) = uv - int(vdu)\n= xln(x) - int(x(1/x)dx)\n= xln(x) - int(dx)\n= xln(x) - x + C, where C is a constant
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
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
171290
I think you mean "integral" of x^(n-1)* exp(-(log)^2)/2 and I THINK that the e and the log should cancel so you would have X^(n-1) * -X^2/2 = x^n-1+2 /2 so the integral of that would be... X^(n+1)/(2n+4)
832686
Take the integral of this function, like this:\n\n∫ (10 + x²) / x³ =\n\n∫ (10 / x³) + ∫ (x² / x³) =\n\n10 ∫ (1 / x³) + ∫ (1 / x) =\n\n10 (-1 / 2x²) + ln(x) + C = \n\n-5 / x² + ln(x) + C\n\n~ ♥ ~
274471
i made a little comment for you on your integral answer:)
408775
1/(1+e^x)=(1+e^x -e^x)/(1+e^x)=1-  e^x/(1+e^x)=\n1- (1+e^x)'/(1+e^x)\n& u'll integrate
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
737583
A bit tricky to integrate, but is like this (i could mistake!):\n\na)\n9^(1/5)\nS pi ( (3 sqrt(x))^2 - x^3^2) dx\n0\n\nb)\n9^(1/5)\nS pi ( (3 sqrt(x)-6)^2 - (x^3-6)^2) dx\n0\n\nc)\n9^(3/5)\nS pi ( (y^2/9)^2 - (y^(1/3))^2) dy\n0\n\nd)\n9^(3/5)\nS pi ( (y^2/9-4)^2 - (y^(1/3)-4)^2) dy\n0\n\nRegards