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
407911
So you are doing some numerical analysis... if you have a spreadsheet program you can use that. Use the trapezoid rule or Simpson's rule for some small increment in x.  Your spreadsheet program is able to evaluate arctan.  The second question can be evaluated with Newton's method.
890950
Did they tell you that\nIf x is very close to 0, then sin(x)=tan(x)=x;\n(sin(x/2))^2 / x^2 = (x/2)^2/x^2=1/4\n(tan(2x))^3 / x^3 = (2x)^3/x^3=8
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]
796927
try the link below...... hope it can put an end to ur integral problems
1026215
ok i got the answer to this\ni just solved it now and here's the solution.if u have time and patience then u can go through it ........\n\ntan a{tan (a+60)} + tan a {tan (a -60)} + tan (a+60)tan(a-60)\n\n= tan a(tan a + tan60) / (1-tan a tan60) + tan a (tan a - tan 60) / (1+tan a tan60) + {(tan a + tan 60) / (1- tan a tan 60) } {(tan a -tan 60)/1 + tan a tan 60)}\n\nnote ^ is to the power of,in case u didn't know.and one more thing,make sure u know all the formulae used in this problem.this will help u understand my steps better! and sqrt = sqaure root of .\n\n=( tan^2 a+ tan a sqrt3) / (1 - sqrt3tan a) + (tan ^2a - sqrt 3 tan a )/(1 + sqrt 3 tan a ) + (tan ^2a - 3)/(1 - 3 tan^2 a )--- (this particular thing (tan ^2a - 3)/(1 - 3 tan^2 a ) i got it using the formula (a+b)(a-b)=a^2-b^2 after mulipying {(tan a + tan 60) / (1- tan a tan 60) } {(tan a -tan 60)/1 + tan a tan 60)}\n\nand now that becomes\n= {(1 + sqrt3 tan a )(tan ^2 a+ sqrt 3tana) + (1 - sqrt 3 tan a)(tan^2 a - sqrt3 tan a) + tan^2 a -3} / {1 - 3tan^2a}\n=( tan^2a + sqrt3tan^3a + sqrt 3 tana + 3tan^2a + tan^2a - sqrt 3tan ^3 a - sqrt3tan a + 3tan a  + tan^2 a -3) / ((1 - 3 tan^2 a )\n\nnow u can cancel out + sqrt3tan^3a  and - sqrt3tan^3a ,+sqrt 3 tan a and -sqrt 3 tana .\n\nwhich becomes\n= (9tan^2a - 3) / (1 - 3tan^2a)\n= -3(-3tan^2a + 1) / (1 - 3tan^2 a) \n= -3\n\n\nall the best for your studies dear friend.
413333
let suppose that \nr(@) = 4/ sqrt( 4cos^2(@) +sin^2(@) )\nthen x(@(t) ) = r(@) . cos (@) emplies\n\nx^2 (t) = r^2 (@) . cos^2(@)\n\nx^2 = 16 / (4 cos^2 + sin^2)  . cos^2\n\nso x^2(t) = 16/ (4+ tan^2(t) )\n\nBut tan( @(t) ) = y(t) / x(t)\n\nThus: x^2 = 16 / (4+ y^2 / x^2 )\nThus: \nEQ : x^2/4 + y^2/16 =  1\n\nCheck @=0  --> r(@=0) = 2\nAnd (x=2 , y=0 ) is solution of EQ
801525
This is a tricky one. You have to use a couple of steps to get this one. First you use a u-substituion to "simplify" it and then use a trig substitution.\n\nIf you let u=cos(x) then du=-sin(x) then the integral becomes\n&#92;int -1/(1+u^2) du.\n\nNow let u=tan(s) and du=sec^2(s) and the integral is\n&#92;int -sec^2(s)/(1+tan^2(s)) ds = &#92;int -sec^2(s)/sec^2(s) ds = &#92;int -1 ds.\n\nThe integral is -s + C where C is any constant. We now have to undo the steps to get it back in terms of x. First, we undo the trig substituion. Since u=tan(s) then the integral is\n\n-arctan(u)+C\n\nSince u=cos(x) we get\n\n-arctan(cos(x))+C
283258
I=int dx/[(x^2)*(x^2+1)^.5]\n\nLet x=tan u\n   dx=secu^2 du\n\nSubtitute\n\nI=int (secu^2 du)/[(tanu^2)*(tanu^2+1)^.5]\n\nBut tanu^2+1=secu^2\n\nThen \nI=int (secu^2 du)/[(tanu^2)*(secu^2)^.5]\n--> \nI=int (secu^2 du)/[(tanu^2)*(secu)]\n-->\nI=int (secu du)/[(tanu^2)]\n\nBut\nsecu=1/cosu ..... and tanu=sinu/cosu\n\nThen\nI=int (1/cosu)/(sinu/cosu)^2\nI=int (1/cosu)*(cosu/sinu)^2\nI=int cosu/(sinu)^2\nI=int (cosu)*(sinu)^-2\nI= -(sinu)^-1\nI= -cscu + c .......... c is a constant\n\n\nBut tanu=x --> sinu=x/[(x^2+1)^0.5]\n\nthen\n-cscu= -1/sinu= -1/[x/(x^2+1)^0.5]= -[(x^2+1)^0.5]/x\n\nthen\nI = -[(x^2+1)^0.5] / x + c
86803
0.77(with a little horizontal line right above the last 7)
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
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...
1130920
You can't solve this using traditional techniques since you'll either have a logarithm or an exponential on one of the x variables.  The people claiming Taylor series and techniques like Newton-Raphson approximation are most likely on the right track.  I would use a CAD program, or graph the function y(x)=(1+x)^2-2^x and find the zeroes that way by zooming and such (on at TI-83 or similar).\n\nIn this case, I'll use MathCAD.  The solutions are:\nx={-1.57862, 0, 5.31972} (two of these are approximate, by the way)
149854
babymegan101 is wrong because the derivative of xsin(ax^2+b)+C is sin(ax^2+b) + 2ax^2*cos(ax^2+b). (And we expect the derivative to be sin(ax^2 + b)). I've tried every trick in a physicist's toolbox to solve this integral but I have not succeeded. Therefore I propose that there isn't a definitive answer to the question, and that it can only be solved by approximation using numerical methods.
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.
662965
integral of (tan(x))^(1/3) ... hmmm ...\n\nSubstitute u = tan(x)\nso that du = sec^2(x) dx = (1 + tan^2(x)) dx = (1 + u^2) dx\nand so dx = du / (1 + u^2)\n\nThe substitution therefore yields the\nintegral of (u^(1/3) / (1 + u^2)) du\n = integral du u^(1/3)(1 - u^2 + u^4 - u^6 + u^8 - ...)\n = integral du (u^(1/3) - u^(7/3) + u^(13/3) - u^(19/3) + ...)\n\n = (3/4) u^(4/3) - (3/10) u^(10/3) + (3/16) u^(16/3) - ... + C\n\n = 3 u^(1/3) ( u/4 - (u^3)/10 + (u^5)/16 - ... )\n\nwhich can be written as\n\n3 u^(1/3) sum{ n = 0 to infinity of (-1)^n u^(2n+1) / (6n+4) }\n\nthen do the "+ C" thing and replace u by tan(x)\n\nI would estimate -- an educated guess -- that the radius of convergence of u = tan(x) is 1.
933478
Using the Wolfram Integrator:\nhttp://integrals.wolfram.com/index.jsp\n\nOne obtains\n1/10 sec(2x)tan^4(2x) + 2/15 sec(2x)tan^2(2x) + 4/15 tan(2x)\n\nUsing the other poster's logic - you get (u^2 + 1)^3
161615
Assume n>2 otherwise problem is simple\nI will use "Int" for integral sign\n\nA(n)= Int (tan ^n x) dx = Int (tan^ (n-2) x tan^2 x dx)\n\ntan^2 x = sec^2 x -1\nA(n) = Int tan^(n-2) x sec^2 x dx - Int tan^(n-2) x dx\n\nA(n) = A1(n) - A(n-2)\n\nA1(n) = Int tan^(n-2) x sec^2 x dx\nPut tan(x)=t\nsec^2 x dx = dt\n\nA1(n) = Int t^(n-2) dt = t^(n-1)/(n-1) \n= (tan x)^(n-1)/(n-1)\nThus we have the relation\n\nA(n) = (tan x)^(n-1)/(n-1) - A(n-2)\n\nEvaluate \n\nA(1) = Int tan x dx = -ln cos x +c\nA(2) = Int tan^2 x dx = Int (sec^2 x - 1)dx = tan x -x\n\nNow continue with the rucursion given above for n>2