What is the antiderivative of (sin 2x)e^cos2x?
971343
neither\nit is -1/2 e^cos 2x + C\n\nto confirm differentiate the result and find itr\n\nif you need to integate let cos 2x = y\nthen -2 sin 2x dx = dy\nsin 2x dx = -dy/2\n\nput the value we need to integrate - e^y dy/2 \nresult = - e^y/2 + C\n       = -1/2e^(cos 2x) + C
484958
sin 2x=2 sinx . cos x\ncos 2x=(cos x)^2-(sin x)^2\ncos 2x=2 (cos x)^2 -1\ncos 2x=1- 2 (sin x)^2\ntan 2x=(sin 2x)/(cos 2x)\n\nor you can \nsin 2x=1\nlet 2x=y\nsin y=1\ny=90 degree + 360 degree\n2x=90 degree + 360 degree\nx=45 degree + 180 degree\nso x={45,225,....}degree\n\ntan x+cot x=2/sin 2x:\ntan x+cot x\n=(sin x/cos x )+(cos x/sin x)\n=((sin x)^2+(cos x)^2)/(sin x . cos x)\n=1/(sin x . cos x)\n=2/2(sin x . cos x)\n=2/sin 2x\nso it is an identity
714113
secant x              = 1 / cos x\nderivative             = [cos x (0) - 1 (-sin x)] / cos2 x     (I mean cos square)\n                           =  sinx / cos2 x\n                           =  tan x / cos x                (sin x / cos x = tan x)\n                           = tan x sec x                   (1 / cos x = sec x)\n\nDone! Hope this can help.
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
408737
Need to use the product rule here:\n\nd/dx(1/2*tanxsinx)\n\n= [d/dx(1/2tanx)]sinx + 1/2tanx*d/dx(sinx)\n= (1/2sec^2x)sinx + 1/2tanx*cosx\n= sinx/2cos^2x + sinx/2cos^2x\n= tanx/cosx
749606
L.H.S. (tanx - cotx)/(tanx + cotx) + 2cos^2x = (sinx/cosx - cosx/sinx)/(sinx/cosx + cosx/sinx) + 2cos^2x = (sin^2x - cos^2x)/(sin^2x + cos^2x) + 2cos^2x = (sin^2x - cos^2x)/1 + 2cos^2x = sin^2x - cos^2x+ 2cos^2x = sin^2x + cos^2x = 1 = R.H.S.
831528
"What i know: it can also be written as (x^2 + cos x) ^ -1/2\nand the derivitive of that = (2x - sin x) ^ -1/2"\n\nWhat you said here was to take the derivative of the inside function, but what you do is take the derivative of the outside function and multiply it by the derivative of the inside function.\n\nIf you have a composition of two functions such as h(x) = f(g(x)), then h'(x) = f'(g(x)) * g'(x) [or as in your notes, f(g) * g']\n\nThat is why you must use the chain rule for 1 and 3 but not for 2, since 2 does not involve a composition:\n\n1. f(x) = (x^2 + cos x) ^ (-1/2)\n\nf(x) = g(h(x)) where g(x) = x^ -1/2 and h(x) = x^2 + cos x\n\n3.  f(x) = 2 ^ (x^2 - sin x)\n\nThis isn't a product- it's an exponential function of another function.\n\nf(x) = g(h(x)) where g(x) = 2 ^ x and h(x) = x^2 - sin x
10077
e^(ix)=cos(x)+i*sin(x)\n\nx^i=e(i*ln(x))\n\nx^i=cos(ln(x))+i*sin(ln(x))
532184
One method is to integrate by parts twice. There is another way where you can use complex methods: \ncos2x = Re e^(2ix)\n[Re means the real part]\n\nIntegral e^(-x)cos2x dx\n= Integral e^(-x) Re e^(2ix) dx\n= Re Integral e^((2i-1)x) dx\n= Re e^((2i-1)x)) / (2i-1) + C\n= Re e^(-x) e^(2ix) / (2i-1) + C\n\nWrite the above in the standard A + iB and then you can find your answer. Good luck.
511465
sin x = (1/2i)(e^ix-e^-ix)\ncos x = (1/2)(e^ix+e^-ix)\n\nsolutions of a simple harmonic oscilator are\ny = A sin x + B cos x or\ny = C e^ix + D e^-ix
962855
should be e^u*du/dx\n\nd[e^(x^2)] = e(x^2)*2*x
909136
Let I = ∫e^s sin(t - s) ds\n= e^s sin(t - s) - ∫e^s *( -1) * cos(t - s)ds\n= e^s sin(t - s) + ∫e^s cos(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s * (-1) * -sin(t - s)ds\n= e^s sin(t - s) + e^s cos(t - s) - ∫e^s sin(t - s) ds\n= e^s sin(t - s) + e^s cos(t - s) - I\nSo 2I = e^s (sin(t - s) + cos(t - s)\nThus I = ½(e^s (sin(t - s) + cos(t - s)) + K\nSo ∫[s = 0 to s = t] e^s sin(t - s) ds\n= ½[e^s (sin(t - s) + cos(t - s)] from s= 0 to s = t\n= ½[(e^t (sin(t - t) + cos(t - t)) - (e^0 (sin(t - 0) + cos(t - 0))]\n= ½[(e^t(1) - 1(sint + cost)]\n= ½(e^t - sint - cost)
953609
2cos(x) - sin(2x) = 0 \nremembering that sin(2x) = 2sin(x)cos(x), \n2cos(x) -2sin(x)cos(x) = 0\n2cos(x) * [1 - sin(x)] = 0\nor cos(x) = 0 or 1-sin(x) = 0\ncos(x) = 0 or sin(x) = 1\n\nπ/2, 3π/2
889431
1. Find the antiderivative of 2x + 2sin(x).\nIt is f'(x) = x^2 - 2cos(x) + C\nSince f'(0) = 2, then you can plug in 0 for x and set this equal to 2.\n0^2 - 2cos(0) + C = 2\n-2 + C = 2\nC = 4\nf'(x) = x^2 - 2cosx + 4\nNow find this antiderivative.\nf(x) = (1/3)x^3 - 2sinx + 4x + C\nSince f(0) = 2, we have\n(1/3)0^3 - 2sin(0) + 4(0) + C = 2\nC = 2\nSo f(x) = (1/3)x^3 - 2sin(x) + 4x + 2\nTo find f(2) just replace x with 2 and evaluate.\n2.  Find the antiderivative of 7x +4\nIt is f(x) = (7/2)x^2 + 4x + C\nSince it passes through the point (9,5) that means x = 9 and f(x) = 5\n5 = (7/2)9^2 + 4(9) + C\n5 = (7/2)81 + 36 + C\n5 = 567/2 + 36 + C\n5 = 639/2 + C\nC = 5 - 639/2 = -629/2\nThus, f(x) = (7/2)x^2 + 4x - 629/2\nPlug in x = 3 and solve for f(x).
195729
you dirty dirty man... making me derive... *dusts off brain* alright, give this a shot...\n\nstarting with the inner terms....\n\nderivative of x... 1\n\nderivative of 4sin(x)... 4cos(x)\n\nderivative of 2... 0\n\nderivative of the larger term 1/(2(4sin(x)+2)^(1/2))\n{term needs to be taken to the bottom of the fraction as the exponent is negative}\n\nCompiled we get... 4cos(x)/(2(4sin(x)+2)^(1/2))\n\ninserting the 0... we get... 4/(2(2)^(1/2))... or 2/2^(1/2)\n\nrationalize... multiply by 2^(1/2)/2^(1/2) and we end up with....\n\n2^(1/2)
343985
need to use the chain rule i.e.\n\nf'(pq) = p'*q + p*q'\n\nwe get:\nae^ax/(sin(bx+c)) - be^ax*cos(bx+c)/[(sin(bx+c))^2]\n\nThis is the correct answer\n\n*edit* srry couldnt see all of it if its on one line so split over two lines the answer is:\n\nae^ax/(sin(bx+c)) \n-be^ax*cos(bx+c) / [(sin(bx+c))^2]\n\n\nand ofcourse if you like, further simplification is possible with cos(bx+c)/sin(bx+c) =1/tan(bx+c)=cot(bx+c), also you can take a factor of e^ax out if you like but the above answer is perfectly accebtable.
906876
2cos^2x-1+1/2sinxcosx\n=2cos^2x/2sinxcosx\n=cosx/sinx\n=cotx