integrate 1+tan^2 x / 1-tan^2 x = ???? and simplified the answer to 1/2 In(1+tan x / 1-tan x) + c?
114706
Change of variable t=Tan[x]\nwe have dx=(1+Tan[x]^2)dt\n\nSubstitute:\nIntegral[(1+Tan[x]^2)\n/(1-Tan[x]^2)]dx\n=\nIntegral[dt/(1-t^2)]\n=...\n\nusing partial fraction decomposiotion\n2/(1-t^2)=1/(t+1)-1/(t-1)\n\n...=\n(1/2)(Integral[dt/(t+1)]\n-Integral[dt/(t-1)])\n=\n(1/2)(Ln[Abs[t+1]]\n-Ln[Abs[t-1]])\n=\n(1/2)Ln[Abs[(t+1)/(t-1)]]\n\nreplacing back t we have\n\n(1/2)Ln[Abs[(Tan[t]+1)\n/(Tan[t]-1)]]\nplus C\n\nNote that Abs is needed!\nIf it is used then we can write the result as you requested:\n\n(1/2)Ln[Abs[(1+Tan[t])\n/(1-Tan[t])]]\nplus C
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.
854468
Induri Kartic is almost right but she has missed a certain steps.\n\nlet x = tan A , y= tan B z = tan C (A B C are acute angles) \n\nnow tan(A+B+C) = tan (A+D) using B+C = D\n= (tan A + tan D)/(1- tan A tan D)\n= tan A + (tan B+tan C)/(1- tan B tan C)/denominator say a\n\nnumerator = tan A + tan B + tan C - tan A tan B tan C / some thing say b\nwe have not ecaluated ab \ntan (A+B+C) = (x+y+z-xyz)/ab = 0\nso evaluation of a b was not required\n\nso A+B+C is  pi (as A+B+C < 3pi/2)\n\nnow tan 3(A+B+C) = 0\nso tan 3A + tan 3B + tan 3C = tan 3A \n\ntan 3A = (3tan A -tan^3 A)/(1-3 tan^2 A) = (3x-x^3)/(1-3x^2)\nputting values we get the resultant immediately
858246
Interpretation # 1\n\n2/3 tan x - 2 tan² x =1\n\nso 6 tan² x - 2 tanx + 3 = 0\n\nDiscriminant < 0 so no real solutions\n\nInterpretation # 2\n\n2 tan (x/3) - 2 tan² x =1\n\ntan 3t = tan (t + 2t)\n= (tan t + tan 2t)/(1 - tan t tan 2t)\n= (tan t + 2tan t/1 - tan² t)/( 1 - tan t*2tan t/1 - tan² t)\n= (tan t (1 - tan² t) + 2tan t)/ (1 - tan² t - 2tan² t)\n= (3tan t - tan³ t)/(1 - 3tan² t)\n\nLet tan x/3 = p Then tan x = p(3 - p²)/(1 - 3p²)\nSo 2 tan (x/3) - 2 tan² x\n= 2p - 2 (p(3 - p²)/(1 - 3p²))²\n\n= [2p(1 - 3p²)² - 2p²(3 - p²)²]/(1 - 3p²)² = 1\nSo 2p[(1 - 3p²)² - p(3 - p²)²] = (1 - 3p²)²\nie (2p - 1)(1 - 3p²)² - 2p²(3 - p²)² = 0\nSo (2p - 1)(1 - 6p² + 9p^4) - 2p²(9 - 6p² + p^4) = 0\n\nSo 18p^5 - 9p^4 - 12p³  + 6p² + 2p - 1 - 2p^6 + 12p^4 - 18p² = 0\nie 2p^6 - 18p^5 - 3p^4 + 12p³  + 12p² - 2p + 1 = 0\n\ntan(x/3) ≈ 1.09073, 9.08455 (by equation solver)\nSince 0 ≤ x < 2π then 0 ≤ x/3 < 2π/3\n\nSo x/3 ≈ 0.82877, 1.46116\nSo x ≈ 2.4863, 4.3835
885632
Yes you are correct.\n\nx=tan t = (sin t)/(cos t)\nso cos t = (sin t)/x\nSo y^2 = (sin^2 t)/x^2 = (1 - cos^2 t)/x^2 = (1 - y)/x^2\n\nRearranging this you get y = 1/(x^2 +1)
983394
Depends\n\nlim θ→π/2- (ie from below) tanθ = +∞\n\nlim θ→π/2+ (ie from above) tanθ = -∞ \n\nThis is because in the 1st quadrant tanθ >0 and in the second quadrant tanθ <0
743452
Assuming that x^4 is not in the denominator:\n\n(x-2/x^2) simplifies to  (x^-1) - (2x^-2)\n\nTherefore, the problem becomes:\n\nint[ x^-1  -  2x^-2  + x^4 ]\n\nNow you can use the power rule to integrate each term:\n\nln(x) + 2x^-1 + (x^5)/5  +  C
874788
cos ^-1 0 = pi/2\n1/2 cos^-1 0 = pi/4\n\nso tan(1/2 cos^-1(0)) = tan pi/4 = 1
969104
okay,\n\ncos(x-y)      cos(x)*cos(y) + sin(x)*sin(y)\n----------  =   ----------------------------------- \ncos(x+y)     cos(x)*cos(y) - sin(x)*sin(y)\n\n(multiply and divide the fraction by cos(x)*cos(y))\n\n     1 + tan(x)*tan(y)\n=   ---------------------\n     1 -  tan(x)*tan(y)
317595
If f is an antiderivative of (tanx)^2/(x^2+1) such that f(1)=0.5, then f(0)=__?\n\nnumerical integral of (tan x)^2/(x^2 + 1) from 0 to 1 = 0.345\nf(x) = numerical integral of (tan x)^2/(x^2 + 1) from 0 to x + 0.155\nso f(0) = 0.155
72362
I forgot how to do those, but it's some sort of trig identity isn't it?  Like\n\nsqrt (1-tan [x])\n\nProbably not though.
817617
tan(A-B)+tan(B-C)+tan(C-A)\n= (tanA - tanB)/(1 + tanAtanB) + (tanB - tanC)/(1 + tanBtanC) + (tanC - tanA)/(1 + tanAtanC)\n= [(tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC)]/[(1 + tanAtanB)(1 + tanBtanC)(1 + tanAtanC)]\n= 0\nie (tanA - tanB)(1 + tanBtanC)(1 + tanAtanC) + (tanB - tanC)(1 + tanAtanC)(1 + tanAtanB) + (tanC - tanA)(1 + tanAtanB)(1 + tanBtanC) = 0\nOn expanding\n(tanA - tanB)(1 + tanAtanC + tanBtanC + tanAtanBtan²C) \n+ (tanB - tanC)(1 + tanAtanC + tanAtanB + tan²AtanBtanC)\n+ (tanC - tanA)(1 + tanAtanB + tanBtanC + tanAtan²BtanC)\n\n= tanA - tanB + tan²AtanC - tanAtanBtanC + tanAtanBtanC - tan²BtanC + tan²AtanBtan²C - tanAtan²Btan²C \n+ tanB - tanC + tanAtanBtanC - tanAtan²C + tanAtan²B - tanAtanBtanC + tan²Atan²BtanC - tan²AtanBtan²C\n+ tanC - tanA + tanAtanBtanC - tan²AtanB + tanBtan²C - tanAtanBtanC  + tanAtan²Btan²C - tan²Atan²BtanC\n= tan²AtanC - tan²BtanC - tanAtan²C + tanAtan²B - tan²AtanB + tanBtan²C\n=0 \nDivide all by tanAtanBtanC\nSo tanA/tanB - tanB/tanA - tanC/tanB + tanB/tanC - tanA/tanC - tanC/tanA = 0\n\nie tanA/tanB + tanB/tanC + tanC/tanA = tanB/tanA + tanC/tanB = tanA/tanC\n\nWhence tanA/tanB + tanB/tanC + tanC/tanA = cotA/cotB + cotB/cotC + cotC/cotA\n\nSince a triangle then tan(A + B) = tan(π - C) = - tanC\nSo (tanA + tanB)/(1 - tanAtanB) = -tanC\nie tan A + tan B = tanAtanBtanC - tanC\nso tanA + tanB + tanC = tanAtanBtanC
591905
1+2(tanx)^2+(tanx)^4\n= (1+(tanx)^2)^2\n= ((secx)^2)^2\n= (secx)^4\n= (cosx)^-4
106150
I am not sure what you mean. I assume you mean what is tan(x+a)=sin(x+a)/cos(x+a)\n=[sin(a)cos(x)+sin(x)cos(a)] / [cos(x)cos(a)-sin(x)sin(a)]\n=[tan(a)+tan(x)] / [1-tan(x)tan(a)]\n\nSo tan(x), if the argument x is translated by a, becomes this expression.
235956
tan(nPi/2) when n=7\n\nor tan(nPi/4) when n=2\n\nor if you need one expression,\n\n1/[Tan(nPi/2) Tan(nPi/4)].\n\nRemember, tan = sin/cos...
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
521923
Let 'I' denote the integral sign.\nWe will use the method of substitution.\nLet e^2x = z.\nThen taking differentials of both sides 2*e^2x*dx = dz or 2*z*dx = dz or dx = dz/(2*z).\n\nNow 1/((e^-2x)+( e^2x)) can be written as (e^2x)/[1 + (e^2x)^2]\n\nHence given integral = I[(e^2x)/{1 + (e^2x)^2}]dx\n= I[z/(1 + z^2)]{dz/(2*z)} [substituting z = e^2x]\n= 0.5*I[1/(1 + z^2)]dz\n= 0.5*tan-1(z) + C [tan-1(z) denotes tan inverse z]\nwhere C = arbitrary constant of integration\n= 0.5*tan-1(e^2x) + C