solve: (1) integral of (sin7xcos3x)dx (2) integral of (cos3xcos4x)dx?
913883
(sin7xcos3x)\nmultiply and divide by 2\n2(sin7xcos3x)/2\nuse formula\n2sinAcosB\n=sin(A+B)+sin(A-B)\nit becomes\n(sin10x+sin4x)/2\nnow integrate u get\n[(-cos10x)/10+(-cos4x)/4]/2\n\nfor (2)part\nuse formula\n2cosAcosB\n= cos(A+B)+cos(A-B)
906876
2cos^2x-1+1/2sinxcosx\n=2cos^2x/2sinxcosx\n=cosx/sinx\n=cotx
881174
use formulas for cos(2x) and sin(2x) depending on cos(x) and sin (x) and then use the fact that\nsin(2x) = sqrt(1-cos(2x)^2).\nyou will get a quadratic equation that you solve with quadratic formula
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
255922
Im not sure if you need clarification on how to do this. Ill provide a few details on how to get these answers:\n\n1) ∫sinxdx = - cos(x) + C\n-This one is relatively straightforward, just think "what gives me sin(x) when I take its derivative"\n\n2)∫e3xdx = (Im assuming this means e^(3x))\n         = (e^(3x))/3 + C\n-Just think of the chain rule backwards. If you take the derivative of e^3x you'll have to multiply by 3, so when you take the derivative divide by 3.\n\n3)∫cos3xdx = (Im assuming this is (cos^3)x)\n           = ∫((cos^2)x )(cos(x))\n           = ∫(1-(sin^2)x)((cos(x))\n      Now let u=sin(x)\n             du=cos(x)dx\n           = ∫(1-u^2)du\n           = [u - (u^3)/3]\n     Sub back in for u\n           = sin(x) - (sin^3(x))/3 + C\n-Wow, that one was hard to type out. I hope you can follow it somewhat. The trick is to write cos^3(x) as cos^2(x) and cos(x). Then the identity cos^2(x) + sin^2(x)=1 is used. The u substution allows us to find the integral at the end.\n\n4) Derivative of 3x^4 + 2x^2 + x + 3\n                  = 12x^3 + 4x + 1\n-The derivative is defined by some long equation, but you can use the following fact:\nIf f(x)=x^n\nthen f'(x)=nx^n-1\n\nHopefully you understand most of what I did. Im sure someone else will probably explain a lot of this in much better detail. Like I said, I wasnt sure how much detail you were looking for. \n\nPS Ill tell you site that will help you forever in any integral calculus course: http://integrals.wolfram.com/index.jsp
84808
multiply and divide by (1 - sinx)\n\nyou get (1 - sin^2(x))/(cosx(1-sinx)) that is cos^2(x)/(cosx(1-sinx)) that is cosx/(1-sinx)\n\nnow at the denominator you have a function and at the numerator you have the derivative of that function, save for the sign. \nyour integral is \n-ln(|1-sinx|) + c\n\n---- \n\nproof that we're all saying the same thing:\n- ln |1-sinx| = ln (|1 - sinx|^-1) =\n\n= ln |1/(1-sinx)| ---> divide and multiply again for \n(1 + sinx)  ----> = ln |(1 + sinx)/cos^2(x)|\n\n= ln |(secx + tanx)/cosx| = ln|(secx + tanx)secx| = \n= ln |secx + tanx| + ln|secx|\n\nSo you have many ways to write your answer: you can choose the one which you prefer\n- by elegance\n- by minimization of spread error in calculating it\n- by whatever you like\n\nhave a nice day
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.
732786
Using Trig identity 2sin(x)cos(x) = sin2x\n\n(2sin3x)(cos3x) = sin6x
962689
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
889431
1. Find the antiderivative of 2x + 2sin(x).\nIt is f'(x) = x^2 - 2cos(x) + C\nSince f'(0) = 2, then you can plug in 0 for x and set this equal to 2.\n0^2 - 2cos(0) + C = 2\n-2 + C = 2\nC = 4\nf'(x) = x^2 - 2cosx + 4\nNow find this antiderivative.\nf(x) = (1/3)x^3 - 2sinx + 4x + C\nSince f(0) = 2, we have\n(1/3)0^3 - 2sin(0) + 4(0) + C = 2\nC = 2\nSo f(x) = (1/3)x^3 - 2sin(x) + 4x + 2\nTo find f(2) just replace x with 2 and evaluate.\n2.  Find the antiderivative of 7x +4\nIt is f(x) = (7/2)x^2 + 4x + C\nSince it passes through the point (9,5) that means x = 9 and f(x) = 5\n5 = (7/2)9^2 + 4(9) + C\n5 = (7/2)81 + 36 + C\n5 = 567/2 + 36 + C\n5 = 639/2 + C\nC = 5 - 639/2 = -629/2\nThus, f(x) = (7/2)x^2 + 4x - 629/2\nPlug in x = 3 and solve for f(x).
791534
(2asinx*cosx)^2 + a^2(cos^2x - sin^2x)^2\n=4a^2 sin^2xCos^2x + a^2(cos^4x+sin^4x -2sin^2xCos^2x )\n=2a^2 sin^2xCos^2x + a^2cos^4x+a^2sin^4x\n=a^2(cos^2x + sin^2x)^2\n=a^2
281867
Well I assume you are trying to work out x and I also assume that it's not cos [to the power of] (x^3) but Cos [multiptied by] (x^3)\n\nYou need to integrate to remove the d/dx:\n\nCos integrated is Sin and you then have to divide by the differental of what's in the brackets\n\n=> sin(x^3) / 3x^2
564945
4*sin(x)cos^2(x) \n= 2*{2*sin(x)cos(x)}cos(x) \n= 2*sin(2x)cos(x)\n= sin(3x) + sin(x)\nTherefore, sin(x)cos^2(x) = [sin(3x) + sin(x)]/4\n\n4*sin^2(x)cos(x)\n= 2*{2*sin(x)cos(x)}sin(x)\n= 2*sin(2x)sin(x)\n= cos(x) - cos(3x)\nTherefore, sin^2(x)cos(x) = [cos(x) - cos(3x)]/4\n\nHence the expression inside the integral sign is\n[sin(3x) + sin(x) - cos(x) + cos(3x)]/4\n\nThus the integral becomes\n(1/4)*∫[sin(3x) + sin(x) - cos(x) + cos(3x)]dx\n= (1/4)*[∫sin(3x)dx + ∫sin(x)dx - ∫cos(x)dx + ∫cos(3x)dx]\n= (1/4)*[{-cos(3x)}/3 - cos(x) - sin(x) + {sin(3x)}/3] + C\nwhere C = arbitrary constant of integration\n= (1/4)*[{sin(3x) - 3*sin(x)} - {cos(3x) + cos(x)}]/3 + C\n= (1/12)*[- 4sin^3(x) - 4cos^3(x)] + C\n[N.B. cos(3x)=4cos^3(x)-3cos(x) and sin(3x)=3sin(x)-4sin^3(x)]\n= - [sin^3(x) + cos^3(x)]/3 + C
808989
Let t=sinx\n=> dt = cosx dx\n\nx=pi/2 => t=1\nx=0 => t=0\n\nI=integrate(e^t dt) from 0 to 1\n\n=e^t from 0 to 1\n\n=> I= e^1 - e^0 = e-1\n\nAnswer is I=e-1
734906
the integral of [cos 2x dx] is [1/2 sin 2x + c]
1122517
sin^2+cos^2=1\nsin=1/cosec\nsin=opp side/hypotheses side\ncos=adj side/hypotheses side\nhyp^2=opp^2+adj^2\n\nSin*cosec*cos=cos
851033
(cosx-sinx)^2=cos^2x+sin^2x-2sinxcosx so\nit can be 1-2sinxcosx+cosx+sinx