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1 Jan 2010 21:36:02 IST
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1)(x+2)(x+3)dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") 2)If  then Prove that  3) 4)Let =\int_{1}^{x}\sqrt{2-t^{2}}dt "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") . Then find the real roots of the equation =0 "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") ?
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bye ppl
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1 Jan 2010 22:01:19 IST
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i m in a hurry so just giving answer to 1st one . int ( -3 to -1 ) (x+1)(x+2)(x+3) dx Transformation x+2 =y =int (-1 to 1 ) (y-1) y (y+1)dy = 0
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The time u guys take to find the derivative of a function or for finding the equilibrium constant of a reaction or for finding the angle of dispersion of prism or for standing from ur seat to congratulate our team after their win almost in that time one kid die because of poverty. |
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1 Jan 2010 22:53:59 IST
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"Give me some sunshine,
Give me some rain,
Give me another chance,
I wanna grow up once again"
n.a.
eragon24. |
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1 Jan 2010 22:56:26 IST
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"Give me some sunshine,
Give me some rain,
Give me another chance,
I wanna grow up once again"
n.a.
eragon24. |
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2 Jan 2010 18:19:34 IST
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4. Apply lebnitz theorem, So, u will get f'(x) = ( 2 - x2 )1/2 Since, x2 = f'(x) = ( 2 - x2 )1/2 So, x4 = 2 - x2 or, x4 + x2 - 2 = 0 or, x4 + 2x2 - x2 - 2 = 0 or, (x2 - 1) (x2 + 2) = 0 Now since x is real, so x2 = 1 and other value is neglected therefore, x = + 1 which are the required roots
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The best changes often start as a single, simple thought. Think big, and discover the ways to make your dreams real. |
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2 Jan 2010 18:23:14 IST
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3. Try solving integral with the properties of definite integral. U will get (4096)a11/2 / (3465) = (7/8)(pi)a5 Solving it, u get a = ( 33 (35)1/2 ) / 128 Calculation is a bit complicated in this question. I would try it later if it could be solved by some other method.
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The best changes often start as a single, simple thought. Think big, and discover the ways to make your dreams real. |
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3 Jan 2010 01:06:37 IST
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in 3) try putting x =square root of (2a)*sintheta
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even sky is not the limit
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4 Jan 2010 15:37:33 IST
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thank a ton for helping me out.. can u plz post post d complete solution for 3rd one..thnx.. also few more doubts 5)If  then find  at  6)Prove that  7)Show that  8)
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bye ppl
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4 Jan 2010 18:14:43 IST
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I = integral 0-pi xdx/1+cos^2x .........1eqn I=integral 0-pi (pi-x)dx/1+cos^2(pi-x) = (dx/1+cos^2xpi-x)....2eqn adding 1 and 2 2I = integral 0-pi pidx/1+cos^2x I = pi/2 integral 0-pi dx/1+cos^2x = pi/2(0-pi dx/1+cos^2x +0-pi/2 dx/1+cos^2(pi-x)) =pi integral 0-pi sec^2xdx/ 2+tan^2 x .........put tanx = t = pi integral 0-infinity dt/t^2+2 =pi/root2 tan inverse (t/root2) limits 0- infinity =pi^2/2root2
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the secret of getting ahead is getting started !!!! |
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4 Jan 2010 18:23:43 IST
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5) Differentiate using Newton Leibnitz theorem, f'(x) = 2xcos2x/(1+sin2(x)) f'(pi) = 2(pi) 6) (Not able to post using goiit equation editor so i'll just give you the idea) Let sin(t) = x Then t = sin-1(x) i.e. dt = dx/{1-x2}1/2 Therefore the integral reduces to INTEGRAL (0 to pi/2) {log(sint)dt} Let I = INTEGRAL (0 to pi/2) log(sint)dt i.e. I = INTEGRAL (0 to pi/2) log(cost)dt so 2I = INTEGRAL (0 to pi/2) log(sint.cost)dt = INTEGRAL (0 to pi/2) log(sin2t/2)dt = INTEGRAL (0 to pi/2)logsin2tdt - INTEGRAL(0 to pi/2)log2dt But , INTEGRAL (0 to pi/2) log(sin2t)dt = INTEGRAL (0 to pi)(0.5sinzdz) = I i.e. I = -(pi/2)log(2) 8) Is easy. Use the property INTEGRAL(0 to a) f(x)dx = INTEGRAL (0 to a)f(a-x)dx So the integral reduces to I = INTEGRAL (0 to pi)0.5(pi)dx/{1+cos2x} Now, replace cosx = 1/secx. However this makes the function discontinuous at x=(pi/2) So split the integral as INTEGRAL (0 to pi/2) + INTEGRAL (pi/2 to pi) Now replace cosx = 1/secx, put tanx = t and proceed.
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Finally I reveal myself... |
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4 Jan 2010 22:32:53 IST
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the secret of getting ahead is getting started !!!! |
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4 Jan 2010 22:33:28 IST
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the secret of getting ahead is getting started !!!! |
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4 Jan 2010 23:03:29 IST
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dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") I =
dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") (integral property)
2I = cos(x))dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") since 2sinxcosx =sin2x, /2)dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") = 2I
)dx-\int_{0}^{\pi/2}\log(2)dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
Consider, )dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
Put 2x = t So you have )dt "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
Use property of integrals, )dt +0.5\int_{\pi/2}^{\pi}\ln(\sin(t))dt "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
Now you can easily prove that I1 = I hence I = -(pi/2) (log2) =(pi/2)(log(1/2))
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Finally I reveal myself... |
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4 Jan 2010 23:36:37 IST
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This obviously follows from , dx/\sqrt{(1-x^2)} "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
Let  = t "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") so dt = \int_{0}^{\pi/2}\log\sin(x)dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
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Finally I reveal myself... |
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5 Jan 2010 00:20:09 IST
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dx + \int_{\pi/2}^{\pi}\log\sin(x)dx = \int_{0}^{\pi/2}\log\sin(x)dx + \int_{\pi/2-\pi/2}^{\pi-\pi/2}\log\sin(x+\pi/2)dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
dx + \int_{0}^{\pi/2}\log\cos(x)dx = \int_{0}^{\pi/2}\log\sin(x)dx + \int_{0}^{\pi/2}\log\sin(x)dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
BUT 2\int_{0}^{\pi/2}\log\sin(x)dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
Hence, dx = I "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
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5 Jan 2010 21:24:28 IST
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)\sin(x)dx = I "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
INTEGRATE BY PARTS, EVAULATE THE INDEFINITE INTEGRAL FIRST \sin(x)dx = \log\sin(x)\int \sin(x)dx - \int [\frac{\mathrm{d} \log\sin(x)}{\mathrm{d} x}\int \sin(x)dx]dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
\cos(x) + \int [\cos^2(x)/\sin(x)]dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
\cos(x) + \int [\csc(x) - \sin(x)]dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
\cos(x) + \int \csc(x)dx - \int\sin(x)dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
\cos(x) + \log\tan(x/2) + \cos(x)dx "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") (TYPO HERE, IGNORE 'dx')
NOW APPLY THE LIMITS, \cos(\pi/2) + \log\tan(\pi/2/2) + \cos(\pi/2) = 0 "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") .........(1)
NOW FOR THE LOWER LIMIT, AS THE LOG FUNCTIONS BECOME INFINITE AT X=0, WE HAVE TO CONSIDER THE LIMITING VALUE SO, LOWER LIMIT /\sin^{\cos(x)}(x)) + \cos(0) "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
/\(\sin^{\cos(x)}(x)))+ \cos(0) "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
CONSIDER /\(\sin^{\cos(x)}(x)) "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
EVALUATE THIS LIMIT BY L'HOSPITAL RULE, /\(\sin^{\cos(x)}(x)) =-\log(2) "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
SO THE LOWER LIMIT BECOMES + 1 = -\log(2) + \log(e) = \log(e/2) "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") ................(2)
NOW , -(2) = -\log(e/2) = \log(2/e) "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
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Finally I reveal myself... |
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5 Jan 2010 21:28:18 IST
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 denotes logarithm to the base "e".
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6 Jan 2010 16:25:37 IST
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Impossible becomes possible only if u r awesome :) |
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6 Jan 2010 16:34:31 IST
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The third question will give rise to a hypergeometric function........ so dont worry for that one...
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Impossible becomes possible only if u r awesome :) |
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6 Jan 2010 18:52:45 IST
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For the question 7, I got a much simpler method. Put cosx = t , you have, \log\sin(x)dx = \int_{0}^{1}\log(\sqrt{1-t^2})dt "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
Integrate by parts, Take 1 as the function u, and the log term as function v, 
(differentiation is with respect to "t" and not x )   "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
 "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") (theres a typo here, The signs are interchanged within the log term it's actually sqrt(1+t/1-t)) So  + t\log(\sqrt{1+t})+t\log(\sqrt{1-t}) "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") \log(\sqrt{1+t})+(t-1)\log(\sqrt{1-t}) "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") Limits from 0 to 1, \log(\sqrt{1-t}) = \lim_{x\rightarrow1}(\log(\sqrt{1-t}/(1/[1-t])) "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") (This is a  form converted into 0/0. Evaluate using L'Hospital Rule) You will see that this value comes out to be zero Hence the upper limit value comes out to be \log(\sqrt{1+1})+ \lim_{x\to 1}(t-1)\log(\sqrt{1-t}) = -1 + 2\log(\sqrt{2}) + 0 "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.") Lower limit t=0 = 0 Hence the value of the integral \log\sin(x)dx = \int_{0}^{1}\log(\sqrt{1-t^2})dt = \log(2/e) "This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.")
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