i need a proof for this   ∫ tanx dx = ln |secx| + c ?
969937
'\nu= sec x \ndu = tan x sec x dx\ntan x dx = du/sec x = du/u\ntherefore:\nintegral of tan x dx  = integral of du/u = ln |u| + c = ln| sec 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
733318
I think your identity should be :\ncsc^2(x) / [1 + tan^2(x)] = cot^2(x)\nas you forgot to square the tan(x).\n\nWorking on the 1 + tan^2(x) gives:\n1 + sin^2(x) / cos^2(x)\n= [sin^2(x) + cos^2(x)] / cos^2(x)\n= 1 / cos^2(x)\n\nNow we have : csc^2(x) / [1 / cos^2(x)]\n= [1 / sin^2(x)] / [1 / cos^2(x)]\n= cos^2(x) / sin^2(x)\n= cot^2(x), which was to be shown.
906655
I assume that the last ex you wrote is e^x too, right?\ny' =  e^(e^x) + x [e^(e^x)] (e^x)\ny' = e^(e^x) [1+ x e^x]\n\nY''= e^(e^x)[ xe^x + e^x] + e^(e^x)  e^x [1 + xe^x]\ny''= e^(e^x)[  xe^x + e^x + 1 + xe^x]
77120
It doesn't appear to simply significantly from the original expression.  I was able to get it to the following:\n\n(tan^3-tan)*sec\n\nDo the multipling by csc-1 top-n-bottom and then use the fact that cot^2=csc^2-1.  Bring the cot^2=cos^2/sin^2 to the top to get the above.\nGood luck.
730612
tan x/(1 + tan x) = sin x/(sin x  + cos x)\n\ntan t = sin x/cos x\n\n(sin x/cos x)/[1 + (sin x/cos x)]\n\nMultiply big numerator and big denominator by cos x\n\nsin x/[cos x + sin x]
860568
∫tan^(1/3)x dx\nI can predict that this'll involve the use of partial fractions.\n\nLet tanx=u^3\n\n=>(sec x)^2 dx=3u^2 du\n\nnow, (sec x)^2=1+ (tan x)^2=1+u^6\n\n=>dx=(3u^2)/1+ u^6\n\ntherefore, ∫tan^(1/3)x dx=3∫[u X u^2/(1+u^6)]du\n\n=>3∫[u^3/(1+u^6)]du\nlet u^2=t\n=>(2u) du=dt\n\n=>3∫[u^3/(1+u^6)]du=(3/2)∫[t/(1+t^3)]dt\n\nNow, partial fractions will be used\nwe know that, t/(1+t^3)=t/(1+t)(1+t^2-t)...........since a^3 + b^3=(a+b)(a^2 + b^2 + ab)\n\nlet, t/(1+t)(1+t^2-t)=A/(1+t) + (Bt + C)/(1+t^2-t)\n\n=>t/(1+t)(1+t^2-t)=[A(1+t^2 -t) + (Bt + C)(1+t)]/(1+t)(1+t^2 -t)................taking LCM\n\nequating coefficients of u, u^2, constants on both sides\n=>1=-A+B+C...................equating coefficients of u\n     0= A+B .......................equating coefficients of u^2\n     0= A+C .......................equating coefficients of constant\n=>A=-1/3, B=1/3, C=1/3\n\n=> (3/2)∫[t/(1+t^3)]dt=\n(3/2)∫[(1/3)/(1+t)]dt + (3/2)∫[(1/3)(1+t)/(1+t^2-t)]dt\n\n\n=>(-1/2)log|1+t| +  (1/2∫[(1+t)/(1+t^2-t)]dt\n\n=>(-1/2)log|1+t| +  (1/4)∫[(2+2t)/(1+t^2-t)]dt \n\nwe know, d/dt(1+ t^2 - t)=2t -1\n=>(-1/2)log|1+t| +  (1/4∫[(2+2t+1-1)/(1+t^2-t)]dt \n\n=>(-1/2)log|1+t| +  (1/4)log|1+t^2 -t| + (3/4)∫1/(1+t^2-t)dt\n\nwe know, 1+t^2-t=(t-1/2)^2 +3/4\n\n=>(-1/2)log|1+t| +  (1/4)log|1+t^2 -t| + (3/4)∫1/[(t-1/2)^2 + 3/4]dt\n\n=>(-1/2)log|1+t| +  (1/4)log|1+t^2 -t| + \n(3/4)1/(2/sqrt(3) arctan[2(t-1/2)/sqrt(3)\n\n\nsubstituting original values t=u^2\n\n=>(-1/2)log|1+u^2| +  (1/4)log|1+u^4 -u^2| + \n(3/4)1/(2/sqrt(3) arctan[2(u^2-1/2)/sqrt(3)\n\nsubstituting original values u=(tanx)^1/3\n\n=>(-1/2)log|1+(tanx)^2/3| +  (1/4)log|1+(tanx)^4/3 -(tanx)^2/3| + \n(3/4)1/(2/sqrt(3) arctan[2((tanx)^2/3 - 1/2)/sqrt(3) + C\n\nwhere arctan(x) means tan inverse x\nresults used= \n1)∫1/(1+x)dx= log|1+x| + c\n2)∫1/(a^2+x^2)dx=\n(1/a)arctan(x/a) + C...........'a' is a constant\n\n\nP.S. Extremely sorry for the length of the solution but it is correct and each and every explanation is given.
291468
Oh yes i have cracked it at last....\n\nsolution...\n\nLet x=tan-1 (y) (i mean tan inverse)\nthen tan x = y,and cot x = 1/y (coz tan-1 (x) = cot-1 (1/x))\n\nthen ur given equation's Left hand side part (LHS) will be\n\n(1/y)/(1-y) + (y)/(1-(1/y))\n\n= 1/(y-y2)+ y/(1-(1/y))\n\n= 1/(y-y2) + y2/(y-1)\n\n= y2(y-1) - 1/y(y-1)\n\n= y3/y(y-1)\n\nI hope u understood till this step.\n\nNow we have (a3-b3) = (a-b) (a2+ab+b2)\n\nthen the LHS becomes...\n\n(y-1)(y2+y+1)/y(y-1)\n\n= (y2+y+1)/y\n\n= y+1+(1/y)\n\nsubstitute back y=tanx, 1/y = cot x and leave 1 as it is...\n\nU have LHS = 1 + tan x + cot x...Thats ur RHS !!!
982428
LHS\n= (sin^2B-tan^2B)/(1-sec^2B)\n= (sin^2B-tan^2B)/(-tan^2B)\n= (sin^2B/-tan^2B) + (-tan^2B/-tan^2B)\n[splitting the numerator up]\n= (-cos^2B) + 1\n[cosx= sinx/tanx]\n= 1 - cos^2B\n= sin^2B\n= RHS
167162
I do agree with Anniemc46.\nIf the denominator is 1+(tanhx)^2 then it is easy to do.\n\nI have done and uploaded to my web site.\nPlease check the url I provided in the sourse.\n\nThanking U,\nThuvalpakshi.
942804
I'm no calculus genius or anything but...\n∫dx/(sec^2)(5x)\n∫(cos^2)(5x)dx\n∫(1+cos(2*5x))/2 dx\n∫1/2dx + ∫ cos(10x) dx\nx/2 + sin(10x)/20
301328
Homework.\n\nLook at B=A. You should find something interesting about tr(A^tA).
888343
Use the chain rule:  dy/dx = d/dx[(ln x)^x] = x(ln x)^(x-1)d/dx[ln x]
865054
Let u=ln t, e^u=t, du=dt/t, dt=t du\nThen:\n∫ln² t dt = ∫t u² du = ∫u² e^u du\nIntegrating by parts:\nu² e^u - ∫2u e^u du\nu² e^u - 2u e^u + ∫2e^u du\nu² e^u - 2u e^u + 2e^u + C\nt ln² t - 2t ln t + 2t +C\nAnd we are done.
138355
First get rid of 2x by putting t = 2x\n( Read "eye 1" =  and "eye = " etc... capital letter I doesn't show up well on screen)\nI1 = Integral (sec^3 (t) dt/2)\nFor simplicity I will write Int for integral sign\nI = Int sec^3 t dt = Int sec t sec^2 t dt\nApply product rule\n\nI = sec t * tan t - Int sec t tan^2 t dt\nI = sec t tan t - Int sec t (sec^2 t - 1) dt\nExpand\nI = sec t tan t - Int sec^3 t dt + Int sec t dt\nBut second term is -I hence you have\n2I  = sec t tan t + Int sec t dt\n\nThis is standard form to evaluate using \n\nInt sec t = ln tan (pi/4 + t/2) + c\n I = 1/2 (sec t tan t + ln tan (pi/4 + t/2))+ C\nI1 = 1/2 I
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.
898070
ln(x) + 2ln(4) = ln(7) - ln(x)\nln(x) + ln(4^2) = ln(7) - ln(x)\nln(x*4^2) = ln(7/x)\nln(16x) = ln(7/x)\n16x = 7/x\n16x^2 = 7\nx^2 = 7/16\nx = sqrt(7/16)\nx = sqrt(7)/4