Integral Master's Please Help Me?
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To solve for #1, you need to use trigonometric substitution.  When doing trig substitution, you have to note the form of it.  They can be in one of these forms:\na) 1 - [junk]^2\nb) [junk]^2 -1\nc) 1 + [junk]^2 or [junk]^2 + 1\n\nThe above cases mirror the following trig identities.\na) 1 - (cos(x))^2 = (sin(x))^2\nb) (sec(x))^2 - 1 = (tan(x))^2\nc) 1 + (tan(x))^2 = (sec(x))^2\n\nWhen doing trig substitution, you have to keep this in mind.\n\nThe appropriate trig substitution in #1 is c).  You'll want to let x=1/(sqrt(5)tany) and work from there.  The answer is much too huge to put on here, and my goal was to get you started anyway.\n\nTo solve #2, Note that sqrt(2/x) is the same as sqrt(2)/sqrt(x).\n\nIntegral (sqrt(2/x))dx = Integral(sqrt(2)/sqrt(x))dx\n= sqrt(2) Integral (1/sqrt(x))dx (I just pulled out the constant).\n\nAnd then from here, all you have to do is note that 1/sqrt(x) is the same as 1/(x^(1/2)) which is the same as x^(-1/2).  Now it's the matter of using the reverse power rule.\n\nsqrt(2) * Integral (x^(-1/2)) = sqrt(2) * 2 * x^(1/2) + C
706993
Try reviewing your class notes and looking through your text book. Im sure that the answer you need is there. It might take a bit of thought to figure out how to do these problems! Take the time to figure it out, it will be worth it in the long run!
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Try here:\n\nhttp://edweb.sdsu.edu/courses/ET650_online/MAPPS/GDBasics.html
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you can go to this site, you enter in the problem and it solves it for you....http://integrals.wolfram.com/index.jsp
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My answer assume that you have a good knowledge of integral calculus. If it's not the case, then use a soft like Matlab...\n\nNotations :  I will use SIGMA for the 1 to infinite sum and INTEGRAL for the integral between -T/2 and T/2.\n\nIf you consider a periodic signal x(t) with a T = 1/f0 period, its Fourier development is :\n\nx(t) = a0/2 + SYGMA[k=1 -> infinite]  (ak*cos(2*PI*f0*t)) +  SYGMA[k=1 -> infinite] (bk*sin*(2*PI*f0*t))\n\nwhere f0 = 1/T is the fundamental frequency  of the signal, a0/2 the mean value and ak, bk are the Fourier coefficients in cosinus and sinus.\n\nThe Fourier coefficients are calculated like this :\n\nak = 2/T * INTEGRAL[-T/2 -> T/2] (x(t)*cos(2*PI*k*f0*t) dt)    for k >= 0\n\nbk = 2/T * INTEGRAL[-T/2 -> T/2] (x(t)*sin(2*PI*k*f0*t) dt)     for k >= 1      \n\nAfter developpement, you will find the values  of all a and b coefficients in function of k.\n\nI hope this will help :)
274471
i made a little comment for you on your integral answer:)
887510
Differential equations.\nLinear algebra.\nNumerical analysis.
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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
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http://mathforum.org/library/drmath/sets/high_calculus.html
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Get a book from your library on applied math.\n\nInternate must be full of such applictions.
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Not the answer, but a suggestion.\n\nThis is a college level questionm, so I'm assuming that you are in college. Go to the math department library and find other texts on complex analysis. Look at the chapters that cover the same material. Look at the examples the authors have worked out. You may be surprised to find this problem or one enough like it to give you a clue on how to proceed.\n\nThis is not that tough a problem. N years ago, (where 0 < N  < 50) I could have given you an answer, but since then, the analysis sections of my brain have shrunken.
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The primitive of your integral is:\n\ns*(e^(st))*(1/w)*atan(s/w)\n\nYou have to do a trigonometric substitution to get there. Do not always trust the weird answers you get with complex calculators.
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Differential and Integral Calculus\nTopology of the real line\nLinear algebra\nAbstract algebra\nComplex Analysis\nDifferential Equations
414104
Using an integral,\n\nWords = S 3t(3t^2 + 1)^(-1/2) dt\n\nlet u = 3t^2 + 1\nthen du = 6t dt\n\n= (1/2) S u^(-1/2) du\n= (1/2)(2u^(1/2))\n= u^(1/2)\n= (3t^2 + 1)^(1/2)\n\nEvaluating from 1 to 4,\n= (3(4)^2 + 1)^(1/2) - (3(1)^2 + 1)^(1/2)\n= 49^(1/2) - 4^(1/2)\n= 7 - 2\n= 5\n\nAnswer A.
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Here are some sites that have saved my behind.  :)\n\nhttp://integrals.wolfram.com/index.jsp\n\nhttp://www.calc101.com/webMathematica/integrals.jsp\n\nhttp://mcraefamily.com/MathHelp/CalculusTableOfIntegrals.htm
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integral(x,-1,2) - 2 * integral(|x|,-1,2)\nintegral(x,-1,2) - 2 * ( integral(-x,-1,0) + integral(x,0,2))\n\n3/2 - 2( 1/2 + 2)\n3/2 - 3\n-3/2
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Try "Calculus" - By Gilbert Strang\nThis is available along with Student study guide &\nInstructor's manual!\n\nhttp://ocw.mit.edu/ans7870/textbooks/Strang/strangtext.htm