How do I integrate (1+sinx)/cosx?
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
732786
Using Trig identity 2sin(x)cos(x) = sin2x\n\n(2sin3x)(cos3x) = sin6x
640359
let sinx=u\nthen cosxdx=du\n\nand ∫[(cosx)/(9+sin^2x)]dx=\n\n∫du/(9+u^2)=        \n\n\n ( now use this formula\n∫du/(a^2+u^2)= 1/a arctan(u/a)+C)\n\n\n=1/3 arctan(u/3) +C=\n\n\n=(1/3)arctan(sinx/3)+C
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.
163479
lim {1/sinx-1/x}\nx->0\n\n=lim[{x-sinx}/xsinx]\nx->0\n\n=lim[{1-cosx}/{sinx+xcosx}     {according to D hosp's }\nx->0\n\n=[{1-cos0}/{sin0+cos0}]\n={1-1}/{0+1}\n=0/1\n=0 Ans.\nThanks.
408737
Need to use the product rule here:\n\nd/dx(1/2*tanxsinx)\n\n= [d/dx(1/2tanx)]sinx + 1/2tanx*d/dx(sinx)\n= (1/2sec^2x)sinx + 1/2tanx*cosx\n= sinx/2cos^2x + sinx/2cos^2x\n= tanx/cosx
1021739
Since tanx = sinx/cosx,\n\ntan(A-B) = sin(A-B)/cos(A-B)\n\nNow, write out the expression for each term on the right side of this equation.\n\nWhen you finish that, divide each term by cosAcosB and you will get the result you are seeking.
835259
I'll give you a hint and let you try to figure out the rest yourself (unless someone else tells you the exact answer). =)\n\nUse your trig identities:\n\ntan A = sinA/cosA\n\ncot A = 1/tanA or cot A = cosA/sinA
840169
u= x\ndv=cosx dx\n\ndu =dx\nv= sinx\n\n\nintegral of xcosx dx = xsinx - integral of sinx dx\n= xsinx + cosx +C\n`
797886
Have you learned trig identities yet?  I'd check to see if there was a trig identity you could substitute here.\n\nBTW, I am assuming you meant cos²x rather than cosx².\n\nTry this:  cos²x + sin²x = 1, so substitute 1-sin²x for cos²x:\n\n6cos²x + sinx - 4 = 0\n\n6(1-sin²x) + sinx - 4 = 0\n\n-sin²x + sinx + 2 = 0\n\n(multiply by -1 to make it a little easier)\n\nsin²x - sinx - 2 = 0\n\n(sinx-2)(sinx+1) = 0\n\nSo, sinx = 2 or sinx = -1\n\nSolving for x, this means x = sin^-1(2) or sin^-1(-1)
848587
sin^2x=(1-cos2x)/2\nantiderivative of (1-cos2x)/2\n=1/2x-(1/4)sin2x +c
463304
(sinx)^(-1) - sqrt(1-x^2) + C\nStart by multiplying (1+x)/(1-x) by (1+x) on the top and bottom\n[(1+x)(1+x)]/[(1-x)(1+x)]\nMultiply the denominator out but not the numerator.\n[(1+x)(1+x)]/[(1-x)(1+x)] = [(1+x)(1+x)]/(1-x^2)\nNow\nsqrt[(1+x)/(1-x)] \n= sqrt[[(1+x)(1+x)]/(1-x^2)]\n= (1+x)/sqrt(1-x^2)\nNow use trig substitution again letting x = sin(u)\nthen dx = cos(u) du\n(1+x)/sqrt(1-x^2) dx \n= [(1+sin(u))/sqrt(1-(sinu)^2)]cos(u) du\n= [(1+sin(u))/sqrt((cosu)^2)]cos(u) du\n= [(1+sin(u))/cos(u)]cos(u) du \nthe cosines cancel\n= [1 + sin(u)]du\n= u - cos(u) + C\nNow x= sin(u) so u = (sin x)^(-1)\nand using a triangle you can see that cos(u) = sqrt(1-x^2)
546258
cos2x=cos(x+x)=cos^2x-sin^2x=1-sin^2x-sin^2x=1-2sin^2x\ncos2x=1-2sin^2x  \ncos2x+2sin^2x-sinx=1-2sin^2x+2sin^2x-sinx=1-sinx=0\nsinx=1\nx=2kπ+π/2  ,   k=...,-1,0,1,...
248717
Here is what I would suggest. \n\nMultiply the left fraction by sinx/sinx; leave the bottom and do out the top:\n\nsinx + sin^2x + cosxsinx\n--------------------------\n(1+sinx - cosx)(sinx)\n\n\nReplace sin^2x with 1 - cos^2x in numerator\n\nsinx + 1 - cos^2x + cosxsinx\n\nSubtract then add cosx in numerator\n\nsinx + 1 - cosx + cosx - cos^2x + cosxsinx\n\nRearrange (not really necessary but it helped me see the next step)\n\n1 + sinx - cosx + cosx + sinxcosx - cos^2x\n\nFactor cosx from the last 3 terms:\n\n(1 + sinx - cosx) + cosx(1 + sinx - cosx)\n\nFactor this and put it back over the denominator:\n\n(1 + cosx)(1 + sinx - cosx)\n----------------------------\n(1 + sinx - cosx)(sinx)\n\nCancel and get:\n\n(1 + cosx)\n-----------\n  sinx\n\n\nWhew!
847056
dy/dx=y² sin x\ndy/y^2=sinxdx\n\nintegrate\n-1/y = -cosx +K\n1/y  = cosx -K\nuse initial conditions\n1=1-K\nK=0\ny= secx\n\nremember, int{sin}= -cos\n\ni hope that this helps
839872
Okay lets try this one ;\n\n2sin²  – 3sinx  + 1 →\n( sin x – 1)(2sin x – 1) = 0\nSin x – 1 = 0  →  sinx = 1 → x =∏ /2 \nX = 2∏ +∏ /2 , …., x = 2k∏ +∏/2 , ( k Є Z ) \n2sin x – 1 = 0  → sinx = 1/2 → x = ∏/6\nAs sin (∏ -∏/6) = sin ∏/6 = 1/2 \nSo ; \nX = 2∏ + ∏/6 , x = 4∏ + ∏/6 , …., x = 2k∏ + ∏/6 , ( k Є Z) \nX = 2∏ + 5∏/6 , x = 4∏ + 5∏/6 ,….., x = 2k∏ + 5∏/6 , ( k Є Z)\n\nGood Luck dear
730722
put sinx=t\ncosxdx=dt\nI=fcos(pi t)dt with limits -1,+1\nI=(sin(pi t))/pi with limits +1,-1\nI=(sin(pi)-sin(-pi))/pi=0\nthanks & good luck