so u have an integral 0 on bottom to .4 on top of [ tan^-1(x^7) + sin(x^7) ] dx. Use the series to approximate
352131
For the second part of your question, you need to differentiate 1/(1+x)^4 for n times and divide by n! and then substitute x=0 to work out the nth coefficient of the power series. The first constant term is found by substituting 0 into 1/(1+x)^4, so it is 1. The second term is found by differentiating 1/(1+x)^4 once and substitute x=0 and then divide by "1!". The third term is produced by differentiate  1/(1+x)^4 twice and substitute x=0 and then divide by "2!" factorial. You can get as many terms as you want, the series looks someting like:\n\n1 -4x + (20/2!)*x^2 -(120/3!)*x^3 .................\n\nSubstitute x= 0.16 to approximate the number, for 7 terms the answer is close to 0.55223. I might have made mistakes in my calc so you might need to calculate these values again.
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
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!
742785
as x ->pi/4 (1-tan x) ->0 and sec x is not infinite(sqrt(2))\n\nso limit is product =0
957759
a. 0 = 7-x^2, x^2=7, x = +/-sqrt(7)\nArea = Integral[-sqrt(7), sqrt(7)] (7-x^2)dx = 7x - (1/3)x^3 |-sqrt(7), sqrt(7) = 2[7sqrt(7) - (1/3)sqrt(7)^3] ~= 24.69 u^2\n\nb. 0 = x^4-8, x^4 = 8, x = +/-8^(1/4)\nArea = (1/5)x^5-8x |-8^(1/4), 8^(1/4) = 2[(1/5)8^(5/4)-8(8)^(1/4)] ~absv= 21.53 u^2
986397
dear u are integrating a derivative. so the answer is\ne^(arctan x) 0 through 1\n= e^(π/4) - e^0\n= e^(π/4) - 1
712410
limit theta tending to 0 sintheta/theta =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
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]
853051
We have to find derivative of, g(x) = the integral (with lower limit tanx and upper limit x^2) of 1 / [sqrt(2+t^4)] dt.\n\nNow, the derivative of the integrand f(t)=1 / [sqrt(2+t^4)] is \nf'(t)= -2*t^3 / (2+t^4)^(3/2) [By Quotient Rule]\n\nd(g(x)) /dt= d {int(tan x)(x^2) {1 / [sqrt(2+t^4)] }dt} /dt\n=>dy /dt= d {f(t)}/dt (at tan x) - d {f(t)}/dt (at x^2)\n=>dy /dt= f'(t) (at tan x) - f(t) (at x^2)\n=>dy /dt= [2*t^3 / (2+t^4)^(3/2)] (at tan x) - [2*t^3 / (2+t^4)^(3/2)] (at x^2)\n=>dy /dt= [2*(tan x) ^3 / (2+(tan x)^4)^(3/2)]  - [2*(x^2)^3 / (2+(x^2)^4)^(3/2)] \n=>dy /dt= [2*(tan x) ^3 / (2+(tan x)^4)^(3/2)]  - [2*(x^8) / (2+(x^16))^(3/2)]\n\nI hope that answers your question.
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.
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
611207
The only way to do this is by using the reduction formula for int(sin(x)^n), which is valid for n>0.  Unfortuantely, it reduces to another integral for sin(x)^(n-2), for which the reduction formula no longer is valid.  Hence, the only way to do this integral is by using numerical integration techniques.\n\nProbably the best way to do this problem is to convert sin(x)^(1/2) to a taylor series.  \n\nI did this and got:\nx^0.5 - x^2.5/12 + x^4.5/1440 - x^6.5/24192 - 67/29030400*x^8.5 .....\n\nIntegrate:\n\n(2/3)*x^(3/2)-(1/42)*x^(7/2)+(1/7920)*x^(11/2)-(1/181440)*x^(15/2)-(67/275788800)*x^(19/2)......\n\nNeedless to say, after the first 4 terms, any error from a direct integration of sqrt(sin(x) is less than 10^-6, which I would say is pretty damn close.  \n\nThere ya go...
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.
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.
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)
820698
Where does tan x have vertical asymptotes?\nAt -pi/2, pi/2, 3*pi/2, 5*pi/2 etc.\n\nSo this graph you are dealing with must have asymptotes where\n2x - pi/4 = -pi/2, pi/2, 3*pi/2 etc.\n\nAdd pi/4 and divide by 2, and you should find that only two of the results are within the domain (0 to 2pi) you've been given, so they are the asymptotes you want.