Math Integration Master Please Help Me!?
967127
I can see why no-one has posted an answer yet!\nI expect left-hand sum means you multiply 3.5, 8.2, 9.0 and 7.5 in turn by the interval length, which is 0.5, and add together.  The right-hand sum means you do it with 8.2, 9.0, 7.5, and 5.3\n\nplot these figures on a graph, with t on the horizontal axis and g/hr on the vertical axis and try to work out a commonsense answer to the second part.  I don't know off the cuff.
989233
Start by letting I = integral of [sin(x)*sin(nx)]\nUse integration by parts twice.\nLet u = sin(x) and dv = sin(nx)\nAfter using integration by parts twice, you'll end up returning to the original integral you started with, so you will have to solve for I.\n(Your instructor should have shown you this trick)\n\nAnswer: [cos(x)*sin(nx) - n*sin(x)*cos(nx)]/[n^2 - 1]\nI did this quickly, so you should check my answer by working it out.
631241
the integration by part is very long and complicated\ni used it many times\nlet u =logsinx          dv=dx\n   du= cosx/sinx          =x\nxlogsinx-int{xcosx/sinx}\nint{xcosx/sinx}  by sbutitution\nlet u=sinx   x=sin^-1(x)\ndu=cosx dx dx=du/cosx\nyou will get int{sin^-1(u)/u} remmeber u=sinx\nby part\nlet k=1/u          dv=sin^-1(u)\n     -1/u^2        int{sin^-1(u)}by part\n\nint{sin^-1(u)}\nlet m=sin^-1(u)                dv=du\n     =1/sqt(1-u^2)                u du\n=usinx^-1(u)-int{u/spt(1-u^2)}\nint{u/spt(1-u^2)} int this by subitution\n\nthen u will get\nsinxsin^-1(x)+(1/2)(1-sin^2x)\n\nso the final ans\nxlogsinx-sinxsin^-1(x)\n+(1/2)(1-sin^2x)+c\n\nplz try to answer this by yourself ,i hope i gave you the idea,thankx
171039
integrate it by parts : \nint[U.dv] = [U.V] - int[V.du]\nwhere U = x, dv = sin^-1
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
32210
integral of sec(x) = ln(|sec(x) + tan(x)|)\n\n\nsec(x) = (sec(x)(tan(x) + sec(x))/(tan(x) + sec(x))\n\n= (sec(x)tan(x) + sec(x)^2)/(tan(x) + sec(x))\n\nif u = (tan(x) + sec(x)), then du = (sec(x)tan(x) + sec(x)^2) \n\nso the integral is the integral of du/u = ln(|u|)
775488
Ok, Victoria, if you're going to ask for help you have to write the questions correctly.  Problem 1 cannot be solved using algebra the way you wrote it, I doubt you can use complex integration or any other advanced tool to solve that one.  Unfortunately, I just spent the last half hour manipulating it before I remembered that little fact.\n\nPut parentheses in.  Had you put the parentheses in the problem would have been trivial.  Kudos to the helpers who were smarter than I in figuring out the constants were in the exponent.  It's not that hard Victoria, especially when you're asking for help.\n\nWrite the question correctly.\n\n\nPS. Irma, you forgot to multiply the 2 from 4=2^2 through the whole parenthetical statement.  The answer is x=4.\n.
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
704235
I would expand (sin 2x+1)^2, then replace (sin 2x)^2 by\n(1-cos 4x)/2, then integrate term by term.  If you can't do that, I expect another answerer will do the whole thing, or else you can email h_chalker@yahoo.com.au
117158
the integration >>> increase the power by one and divide by the new power (don't forget the the constant)\n\nx^n -------> x^(n+1)/(n+1)\n\ny = 1/(x^2)\ny = x^-2\nthe integration = x^(-1)/(-1) + c\n= -x^-1 + c\n= -1/x + constant
859250
integral = x*tan(x)+ln(sec(x))  (use integration by parts with du = sec^2(x) and v = x).\n\ng'(x) = f(x), so a, c, e and g are true (b is true if f is continuous)\n\nSteve
154380
I posted these sources earlier to someone else who was looking for math help.\n\nGood Luck!!
662213
Anyone can give you the answer, but try here next time you need help with conversion or any other kind of math: \n\n\nhttp://www.gateways2learning.com/BasicMath/BasicMath.htm\n\nGood luck..................
955693
i don't understand the way you\nhave written down the function\n\nsay that you want to integrate\n2x/(2x-1)\nput u=2x-1,>>>x=(u+1)/2,\ndx=du/2,\nint{2x/(2x-1)}dx\n=int{(2*(u+1)/2)/u}*du/2\n=1/2*int{(u+1)/u}du\n=1/2*int{1+1/u}du\n=1/2[u+lnu]+C\nbut u= 2x-1\n=1/2[2x+ln(2x-1)-1]+C\nwhere C is a constant\n\ni hope that this helps
761146
You're trying to equate them.  What you really need to do is take their difference and then take the integral of that term:\n\nA = ∫(lnx - (x-2))dx = ∫lnx dx - ∫(x-2)dx\n\nLooks a little easier now, eh?\n\nThere's still a problem.  To find the limits of integration, you have to solve the 2 eqns simultaneously to get the intersection points.  I don't think there is an algebraic way to do that, so here are the limits (determined numerically); x = .158594 and x = 3.1465
774662
Set u = Arccos(x) such that dx = sinu du.  Then the integral becomes the integral of (u^2)sinu du.  Now the integral of this is (by parts) = -(u^2)sinu + 2*(the integral of) (usinu du).  And the integral of usinu (by parts) = -ucosu+(the integral of)(cosu du).\n\nHope this helps,\n\nSteve
880226
it is not what you specified\n\nIt cannot be solved without identities\n\nsin ^5 t = sin^4 t sin t\n= (1-cos^2 t)^2 sin t\n\nnow let cost = x\nso diffentiate to get - sint dt = dx\n\nso we need to integrate\n\n-(1-x^2)^2 dx\n\n= -dx + 2 x^2 dx - x^4 dx\n\nintegrating we get\n\n-x + 2x^3/3 - x^5/5 + C\n= - cos t + 2 cos^3 t/3 - cos^5 t /5 + C\n\nwhere C is constant of integration