what is the product of integral (Arc tan x)1/2??
913559
0.5 (x) arctan(x) - 0.25 log (x^2+1)
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.
806158
Distribute the -1/2 through the parentheses and you get a \n-1/4 integral e^2x sin x\n\nAdd this to both sides and you will get the 5/4 integral of e^2x sinx
390806
The above theorem about products of integrals is wrong as stated. Counterexample: Int[1]dx = Int[1*1]dx = x + C. But (Int[1]dx)*(Int[1]dx) = x^2 + ax + b.\nInstead, use trigonometric substitution to change square roots into trigonometric functions. Sqrt[25-4x^2] can be written as the leg of a right triangle with hypotenuse 5 and another leg of length 2x. If we let the angle adjacent to the leg of length Sqrt[25-4x^2] be t, then 5cos(t) is the main part of your integrand and x = (5/2)sin(t) implies dx = (5/2)cos(t) dt.
881848
First apply the identities tan x = sin x/cos x and cot x = cos x/sin x:\n\n(tan x - cot x)/(tan x + cot x) + 2 cos^2 x = (sin x/cos x - cos x/sinx)/(sin x/cos x + cos x/sin x) + 2 cos^2 x.\n\nNow get a common denominator and combine some fractions:\n\n(sin x/cos x - cos x/sinx)/(sin x/cos x + cos x/sin x) + 2 cos^2 x =\n((sin^2 x - cos^2 x)/ sin x cos x)/((sin^2 x + cos^2 x)/ sin x cos x) + 2 cos^2 x\n\nWe can cancel the sin x cos x's on bottom, and get\n\n((sin^2 x - cos^2 x)/ sin x cos x)/((sin^2 x + cos^2 x)/ sin x cos x) + 2 cos^2 x = (sin^2 x - cos^2 x)/(sin^2 x + cos^2 x) - 2 cos^2 x.\n\nNow, applying the identity sin^2 x + cos^2 x = 1 gives\n\n(sin^2 x - cos^2 x)/(sin^2 x + cos^2 x) - 2 cos^2 x = (sin^2 x - cos^2 x) + 2 cos^2 x.\n\nSimplifying gives\n\n(sin^2 x - cos^2 x) + 2 cos^2 x = sin^2 x + cos^2 x.\n\nAnd finally\n\nsin^2 x + cos^2 x = 1.
697906
let y = arc tan (x)\n\ntan y = x\ndifferentiate\n\nsec^2 y dy/dx = 1\n\nor(1+tan ^2y) dy/dx = 1\n\ndy/dx = 1/(1+tan^2 y) = 1/(1+x^2)
671135
hmmm... I'm getting that the integral of sec(x)^3 is\n  .5sec(x)tan(x)+.5ln|secx+tanx|\n\nsince....\n\nintegral sec (x)^n =  1/(n-1)*sec(x)^(n-2)*tan(x)+\n(n-2)/(n-1)*integral sec(x)^(n-2) \n\nI'm not sure how it could just be ln|sec(x) + tan(x)|\n\n\nit is (sec (x))^3 and not sec(x^3) right? (just thought i'd double check!)
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!
809968
(tan x- cot x)/( tan x + cot x)+ 2cos^2 x=1\n\nSenx / cos x  - cosx/senx\n---------------------------------  + 2cos^2x\nSenx / cos x  + cosx/senx\n\n(sen^2 x - cos^2x) / senx cosx\n--------------------------------------  + 2cos^2 x\n(sen^2 x + cos^2x) / senx cosx\n\nThe fraction reduces on the above:\n\nsen^2 x - cos^2 x                       + 2 cos^2 x\n\nsen^2x + cos^2x = 1\n\nproved!
1161284
answer : (pi^2)/2\n  \n\nvolume=pi*(int(y^2))(-1 - 1)\n          = pi*(int (1/(1+x^2)))(-1 - 1)\n         = pi*(arc tan x)(-1 -1)\n         = pi^2/2
592160
i think the answer wil be (tanX)/(tanY)\n\nSIN(X+Y)=sin x cos y+cos x sin y\n\nsin(x-y)=sin x cos y-cos x sin y\n\n use cos x cos y to divide the numerator and denominator... there is the answer...
734906
the integral of [cos 2x dx] is [1/2 sin 2x + c]
694637
tan(theta)=x/x^3=1/x^2. So theta=arctan(1/x^2)
861347
This is a very, very difficult (if not impossible) integral.  Which should give you a clue that it's a trick question.\n\nYou are supposed to notice that as x -> 0, x -> the lower bound.  The only reason that the integral might not be zero would be if the function (1 - tan(2t))^1/t approached infinity as t approaches zero.  So lets look at the function.\n\nAs t -> 0, (tan(2t)) -> 0.\nTherefore (1 - tan(2t)) -> 1\nWe know that as t -> 0, 1/t -> infinity.\nHowever, 1^x = 1 no matter what x is (even as x -> infinity)\n\nTherefore, the function approaches one as x -> 0.\nTherefore, the integral must be zero.
935281
The general formula is\narctan(x)-arctan(y)=\narctan[(x-y)/(1+xy)].\nIn your case, x=1+sqrt(2), y=1-sqrt(2), so\nx-y=2sqrt(2) and xy=-1. Thus, you get \narctan(2sqrt(2)/0)=pi/2.
766844
1/tanx - 1/x=\n(x-tanx)/(xtanx)=0/0\nso use l'hopital's rule\n(1-sec^2(x))/(tanx+xsec^2(x))=\n(1-1)/(0+0)= 0/0\nso use l'hopital again\n(-2secxsecxtanx)/(sec^2(x)+\nsec^2(x)+x(2secxsecxtanx))=\n0/(2+0)=0
882948
Just make the following substitution:\nr=sqrt(x^2+y^2)\ntheta=arctan(y/x)\nand then simplify.  Bring everything over to one side and you get your answer.\n\nIf tan(theta)=y/x, then sec(theta)=sqrt(x^2+y^2)/x.\n\nThe problem is a little ugly, but it is doable and then at the end everything ends up canceling nicely.