integrate sin(ln(t)) dt ...?
828677
∫sin(ln(t)) dt\n\n= ∫sin(ln(t)) d(t)/dt dt\n\n= tsin(ln(t)) - ∫t * 1/t cos(ln(t)) dt \n\n= tsin(ln(t)) - ∫ cos(ln(t)) dt \n\n= tsin(ln(t)) - ∫ cos(ln(t)) d(t)/dt dt \n\n= tsin(ln(t)) - (tcos(ln(t)) -∫ t * 1/t *-sin(ln(t)) dt)\n\n= tsin(ln(t)) - tcos(ln(t)) -∫sin(ln(t)) dt\n\nSo 2∫sin(ln(t)) dt = tsin(ln(t)) - tcos(ln(t))\n\nSo ∫sin(ln(t)) dt = t/2(sin(ln(t)) - cos(ln(t)))
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)
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
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?
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`
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.
732220
dx/dt=1/2\ndy/dt=2t\ndy/dx=dy/dt/dx/dt\n=2t/1/2=4t\nat t=-1 slope=-4\nat t=1 slope=4
774662
Set u = Arccos(x) such that dx = sinu du.  Then the integral becomes the integral of (u^2)sinu du.  Now the integral of this is (by parts) = -(u^2)sinu + 2*(the integral of) (usinu du).  And the integral of usinu (by parts) = -ucosu+(the integral of)(cosu du).\n\nHope this helps,\n\nSteve
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
496092
let x equal (tant)^2=>dx=(sect)^2dt. then simply the integrand, u will have something like integral of (cost)(sint)^2 dt. let u be sint. then u know it :D
976124
dy/dx = (dy/dt)/(dx/dt)\ndy/dt = 1/(2√t)\ndx/dt = 1\n\ndy/dx = 1/(2√t)\nx=t\ny=√t\n\nSubstituting t=1/4:\n\ndy/dx = 1\nx = 1/4\ny = 1/2\n\nUsing point-slope form:\n\ny-1/2 = 1(x-1/4)\n\nSimplifying:\n\ny=x+1/4
452294
Okay, this is how I read this problem:\n\nThe integral of 1/(t-1) dt on the interval from 0 to (1-3x+x^2).\n\nFirst look at what you're integrating 1/(t-1) dt.  This can be rewritten (t-1)^-1 dt.  If you substitute u for t-1 and du for dt (the derivative of t-1 = 1) then you have the integral of u^-1 du.  This is ln |u| (ln is the natural log).  Sub t-1 for u and you get ln |t-1|.\n\nNow you use your interval and you get:\n\nln |(1-3x+x^2) - 1| - ln |0|\n\nSimplify and you get:\n\nln |x^2 - 3x| - ln |0|
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))
719088
Substitute u=pi/2-x. Then tan(x)=cot(u), du=-dx and the limits of integration are reversed. So, if we let I_n be the value of the integral, we also have\nI_n is the integral of 1/(1+cot^n (u))du=\nintegral of tan^n(u)/(1+tan^n (u)) du.\n\nNow replace u by x and add to the original to get that\n2I_n =int (1+tan^n (x))/(1+tan^n(x)) dx=pi/2.\n\nThus I_n=pi/4.
14407
This is an integration-by-parts problem. The answer is:\nx*ln(x)+C\n\nThis is how you calculate it:\n\nIntegral of ln(x)*dx\n\nu= ln(x)\ndu=1/x*dx\ndv=dx\nv=x\n\nSo by parts: uv-integral of vdu\n= ln(x)*x-integral of x*1/x*dx\n=x*ln(x)-integral of 1*dx\n=x*ln(x)-x+c
882357
The derivative of ln|f(x)| is 1/f(x) * d f(x) / dx;  this will be\n\n1/[sec5x + tan5x] * d(sec5x) / dx + d(tan5x) / dx\n\nd(sec5x) / dx = d(1/cos5x) / dx = -1/cos^2(5x) * -5sin5x = 5tanx*secx\n\nd(tan5x) / dx = 5sec^2(5x)\n\n(5tanxsecx + 5sec^2(5x)) / (sec5x + tan5x)\n\nFactor out 5sec5x to get 5sec5x * (tan5x + sec5x)/(sec5x + tan5x) = 5sec5x
68076
god man.  what kind of horrible painful calculus class are you taking?\n\nfirst integrate over dx.  the integral of x^y dx is just a power integral, so you have y*x^(y-1).  the integral of y^x dx is just an exponential, so you have y^x/ln y.\n\nnow integrate y*x^(y-1) + y^x/ln y over dy.  and there, my friend, is where i stop because frankly i can't be arsed to do something so brain-curdling.  actually the first term doesn't look so bad, i think if you integrate it by parts it should come out in one step.  don't really want to think about doing the 2nd one.\n\nyou might consider trying some kind of fancy thing like transforming to polar coordinates or something.  might make it easier.\n\nbtw you should know about this site:\nhttp://integrals.wolfram.com/