differentiate sin^-1(4x(1-4x^2)^1/2) with respect to (1-4x^2)^1/2?
956017
well, to differentiate a the given funtion with respect to                (1-4x^2)^1/2, we assume (1-4x^2)^1/2=y. and the function becomes sin^-1(4y) or 1/sin4y. we have to differentiate this with respect to y.\nso, the function, which is cosec4y, on differentiation gives the result as -cosec4y*cot4y*4\nsubstituting the value of y we get the final result in terms of x as\n-4cosec((1-4x^2)^1/2)*cot((1-4x^2)^1/2)\nthe long string in the cot is (1-4x^2)^1/2 ie., y
831486
Differentiate both sides, we have \n\ndy =  (1/x)dx + (1 + tan^2(y))/tany dy\n\ntherefore\n(1 - (1 + tan^2(y))/tany)dy=(1/x)dx,\n\ni.e.\n\ndy/dx=tany/(x(tany - tan^2- 1)).
898033
cos2x = 1 - 2 (sin x) ^2 = sin x\n\nthen 2 ( sin x)^2 + sin x  - 1 = 0\n or (2sin x - 1 )(sin x + 1) = 0\nsin x = 1/2 , - 1\nx = 30 deg, or - 90 deg
956637
Sorry tried a lot, searched a lot & surfed a lot but found no solution, I think Pascal is right.\n\nP.s- Saket it is ∫e^sinx dx & not ∫(e^x)*sin x dx, if it was that, it would have been trivial.
549061
1/x^2 can be rexpressed as x^(-2). As you might know, the formula for a derivative of x^n is nx^(n-1). The antiderivative of x^n is (x^(n+1))/n+1. In this case, it would be x^-1/-1, or -x^-1. To check this, we take the derivative, which gives us (-1)(-1)x^-2, or x^-2. The answer checks out. \n\nSo, in summary, the antiderivative of x^-2 is -x^-1, with the remainder C.
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
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)
495174
if u meant {4/(q^2+2q+1)}- {q/7(q+1)}\nthe answer is\n\n(28-q^2-q)/{7*(1+q)^2}
722288
(x/2)*(100-x^2)^(1/2)+\n50*arcsin(x/10)\n\nSorry it's on two lines, yahoo cuts it off if I leave it on one. arcsin is the inverse sine function.
916864
Ord(e^(4x)-1 -4x) ~ Ord(e^(4x) > Ord(x^2)\nThen:\n\nlim(e^(4x)-1 -4x)/x^2 {x--> oo} = oo\n\nRemember\n\nln x < x^a < e^x < x^x
321475
*(2x/(x^2-4)) + (5/2-x) - (1/2+x)\n*(2x/(x^2-4)) + (5/2-x)((2+x)/(2+x)) - (1/2+x)((2-x)/(2-x))\n*(2x/(x^2-4)) + (10+5x)/(4-x^2) - (2-x)/(4-x^2)\n*(2x/(x^2-4)) + (8+6x)/(4-x^2)\n\nyou should be able to get it from there
200647
sin^2(x)=0.5*(1-cos(2x)).this can be integrated
738333
(1+sin^2x)dx/\n(1+sin^2x)(1-sin^2x)\n=dx/(1-sin^2x)\ndx/cos^2x\nsec^2xdx\ntanx+C
586893
If its differentiate x * sqrt 1 + x^2 , the answer is\n  1 + 2x ( since sqrt 1 is a constant and d x^2/dx = 2x)\n\n \nIf your question is sqrt1 + x^2 answer is 0 + 2x = 2x
198537
arctan'(x) = 1/(1+x^2), so differentiating using the good old rules, we get the result (using 'sq' for the square root to avoid yahoo cutting the equation short):\n\nsq(a+x^2)-1\n-------------------------------\n(x*sq(a+x^2)(1+(x-sq(a+x^2))^2\n\nMan, that's ugly. :-)
400264
split the integral up\n\nyou have:\n(1-2x^2)/x = 1/x - 2x^2/x = 1/x - 2x\n\nintegrating this gives:\n\nIn(x) - 2x^2/2 = In(x)-x^2
463304
(sinx)^(-1) - sqrt(1-x^2) + C\nStart by multiplying (1+x)/(1-x) by (1+x) on the top and bottom\n[(1+x)(1+x)]/[(1-x)(1+x)]\nMultiply the denominator out but not the numerator.\n[(1+x)(1+x)]/[(1-x)(1+x)] = [(1+x)(1+x)]/(1-x^2)\nNow\nsqrt[(1+x)/(1-x)] \n= sqrt[[(1+x)(1+x)]/(1-x^2)]\n= (1+x)/sqrt(1-x^2)\nNow use trig substitution again letting x = sin(u)\nthen dx = cos(u) du\n(1+x)/sqrt(1-x^2) dx \n= [(1+sin(u))/sqrt(1-(sinu)^2)]cos(u) du\n= [(1+sin(u))/sqrt((cosu)^2)]cos(u) du\n= [(1+sin(u))/cos(u)]cos(u) du \nthe cosines cancel\n= [1 + sin(u)]du\n= u - cos(u) + C\nNow x= sin(u) so u = (sin x)^(-1)\nand using a triangle you can see that cos(u) = sqrt(1-x^2)