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kislay (1118)

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the vlaue of 0pi/2  log(sin2x+k2cos2x)dx..is equal to

ans..
pi*log(1+k) - pi*log2

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edison (4934)

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Auume I= 0pi/2  log(sin2x+k2cos2x)dx
 
Also replace x by (pi/2 - x) in above integral, the integral will be I only
add the two integrals to get 2I, now solve to get the desired results

The Scientist does not study nature because it is useful; he studies it because he delights in it, & he delights in it because it is beautiful. If nature were not beautiful, it would not be worth knowing, life would not be worth living. Ofcourse I do not here speak of that beauty that strikes the senses, the beauty of qualities & appearances; not that I undervalue such beauty, far from it, but it has nothing to do with science; I mean that profounder beauty which comes from the harmoniuos order of the parts, & which a pure intelligence can grasp.
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edison (4934)

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Definite Integral
 
A definite integral is an integral
 int_a^bf(x)dx
(1)
with upper and lower limits. If x is restricted to lie on the real line, the definite integral is known as a Riemann integral (which is the usual definition encountered in elementary textbooks). However, a general definite integral is taken in the complex plane, resulting in the contour integral
 int_a^bf(z)dz,
(2)
with a, b, and z in general being complex numbers and the path of integration from a to b known as a contour.
The first fundamental theorem of calculus allows definite integrals to be computed in terms of indefinite integrals, since if F is the indefinite integral for a continuous function f(x), then
 
 int_a^bf(x)dx=F(b)-F(a).
 
(3)
This result, while taught early in elementary calculus courses, is actually a very deep result connecting the purely algebraic indefinite integral and the purely analytic (or geometric) definite integral.
The question of which definite integrals can be expressed in terms of elementary functions is not susceptible to any established theory. In fact, the problem belongs to transcendence theory, which appears to be "infinitely hard." For example, there are definite integrals that are equal to the Euler-Mascheroni constant gamma. However, the problem of deciding whether gamma can be expressed in terms of the values at rational values of elementary functions involves the decision as to whether gamma is rational or algebraic, which is not known.
Integration rules of definite integration include
 
 int_a^af(x)dx=0
(4)
and
 int_a^bf(x)dx=-int_b^af(x)dx.
(5)
For c in (a,b),
 int_a^bf(x)dx=int_a^cf(x)dx+int_c^bf(x)dx.
(6)
If g^' is continuous on [a,b] and f is continuous and has an antiderivative on an interval containing the values of g(x) for a<=x<=b, then
 int_a^bf(g(x))g^'(x)dx=int_(g(a))^(g(b))f(u)du.
(7)
 
Consider the definite integral of the form
 
 I(a)=int_0^(pi/2)(dx)/(1+(tanx)^a),
 
(8)
which can be done trivially by taking advantage of the trigonometric identity
 
 tan(1/2pi-x)=cotx.
 
(9)
Letting z=(tanx)^a,
 
I(a)
= int_0^(pi/4)(dx)/(1+z)+int_(pi/4)^(pi/2)(dx)/(1+z)
(10)
= int_0^(pi/4)(dx)/(1+z)+int_0^(pi/4)(dx)/(1+1/z)
(11)
= int_0^(pi/4)(1/(1+z)+1/(1+1/z))dx
(12)
= int_0^(pi/4)dx
(13)
= 1/4pi.
(14)
 
Another example is
 
 int_(-pi)^piln[2cos(1/2x)]dx=0,
(15)
 
which is nontrivially equal to 0.
 
also
 
int_0^1(t^2lntdt)/((t^2-1)(t^4+1)) = (pi^2(2-sqrt(2)))/(32)
(16)
= 0.18067...
(17)
int_0^(pi/4)(t^2dt)/(sin^2t) = K+1/4piln2-1/(16)pi^2
(18)
= 0.84351...
(19)
int_0^pi(xsinxdx)/(1+cos^2x) = 1/4pi^2
(20)
= 2.467401...
(21)
 
 
int_0^(pi/4)(t^3dt)/(sin^2t) = 3/4piK-1/(64)pi^3+3/(32)pi^2ln2-(105)/(64)zeta(3)
(22)
(23)
= 0.3429474...
(24)
int_0^(pi/2)(x^2)/(x^2+ln^2(2cosx))dx = 1/8pi[1-gamma+ln(2pi)]
(25)
= 0.887759656...
(26)
 
 
 
 C(a)=int_0^1(tan^(-1)(sqrt(x^2+a^2)))/(sqrt(x^2+a^2)(x^2+1))dx,
(27)
which have the special values
C(0) = 1/8piln2+1/2K
(28)
C(1) = 1/4pi-1/2pisqrt(2)+3/2sqrt(2)tan^(-1)(sqrt(2))
(29)
C(sqrt(2)) = 5/(96)pi^2
(30)
(Bailey et al. 2006, pp. 42 and 60).
An amazing integral determined empirically is
 2/(sqrt(3))int_0^1(ln^6xtan^(-1)((xsqrt(2))/(x-2)))/(x+1)dx=1/(81648)[-229635L_3(8)+29852550L_3(7)ln3-1632960L_3(6)pi^2+27760320L_3(5)zeta(3)-275184L_3(4)pi^4+36288000L_3(3)zeta(5)-30008L_3(2)pi^6-57030120L_3(1)zeta(7)],
(31)
where
L_3(s) = sum_(n=1)^(infty)1/((3n-2)^s)-1/((3n-1)^s)
(32)
= 1/(3^s)[zeta(s,1/3)-zeta(s,2/3)]
(33)
 
A complicated-looking definite integral of a rational function with a simple solution is given by
 
 int_0^infty(x^8-4x^6+9x^4-5x^2+1)/(x^(12)-10x^(10)+37x^8-42x^6+26x^4-8x^2+1)dx=1/2pi
(34)
 
 
Another challenging integral is that for the volume of the Reuleaux tetrahedron,
 
 
V = int_0^1[(8sqrt(3))/(1+3t^2)-(16sqrt(2)(3t+1)(4t^2+t+1)^(3/2))/((3t^2+1)(11t^2+2t+3)^2)-(sqrt(2)(249t^2+54t+65))/((11t^2+2t+3)^2)]dt,
(35)
= 1/(12)[3sqrt(2)-49pi+162tan^(-1)(sqrt(2))]
(36)
= 0.4221577...
(37)
 
Integrands that look alike could provide very different results, as illustrated by the beautiful pair
 
int_(-infty)^infty(dx)/((e^x-x+1)^2+pi^2) = 1/2
(38)
int_(-infty)^infty(dx)/((e^x-x)^2+pi^2) = 1/(1+W(1))
(39)
= 0.638103743...
(40)
 
Computer mathematics packages also often return results much more complicated than necessary. An example of this type is provided by the integral
 
 phi(alpha)=int_0^piln(1-2alphacosx+alpha^2)dx=2piln|alpha|
(41)
for alpha in R and |alpha|>1 which follows from a simple application of the Leibniz integral rule.
 
There are a wide range of methods available for numerical integration. The most straightforward numerical integration technique uses the Newton-Cotes formulas (also called quadrature formulas), which approximate a function tabulated at a sequence of regularly spaced intervals by various degree polynomials. If the endpoints are tabulated, then the 2- and 3-point formulas are called the trapezoidal rule and Simpson's rule, respectively. The 5-point formula is called Boole's rule. A generalization of the trapezoidal rule is romberg integration, which can yield accurate results for many fewer function evaluations.
If the analytic form of a function is known (instead of its values merely being tabulated at a fixed number of points), the best numerical method of integration is called Gaussian quadrature. By picking the optimal abscissas at which to compute the function, Gaussian quadrature produces the most accurate approximations possible. However, given the speed of modern computers, the additional complication of the Gaussian quadrature formalism often makes it less desirable than the brute-force method of simply repeatedly calculating twice as many points on a regular grid until convergence is obtained.
 
normal class:
 
 int_0^(2pi)int_0^(pi/4)int_0^4(rhocosphi)rho^2sinphidrhodphidtheta=32pi.
(42)
 
The integral corresponds to integration over a spherical cone with opening angle pi/2 and radius 4. However, it's not clear what the integrand physically represents (it resembles slightly computation of a moment of inertia, but that would give a factor (rhosinphi)^2 rather than the given rhocosphi).

The Scientist does not study nature because it is useful; he studies it because he delights in it, & he delights in it because it is beautiful. If nature were not beautiful, it would not be worth knowing, life would not be worth living. Ofcourse I do not here speak of that beauty that strikes the senses, the beauty of qualities & appearances; not that I undervalue such beauty, far from it, but it has nothing to do with science; I mean that profounder beauty which comes from the harmoniuos order of the parts, & which a pure intelligence can grasp.
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sboosy (3063)

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the general form of the question u have asked is:
 
[0 ][pi/2]  log( a cosx + b sin2 x )
 
= pi * [ log ((a) + (b)) - log 2] 
 
anyway this itself is derived using the principle of partial differenciation in integrals
 
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hsbhatt (5020)

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There's a neat trick you can use in such cases of parametric integration.
 
Call I(k) =  0/2  log(sin2x+k2cos2x) dx
 
Then I'(k) = 0/2  2kcos2x/(sin2x+k2cos2x)  dx [Remember the differentiation is being carried out with respect to k only]
 
I'(k) = 2k 0/2  dx/(tan2x+k2)  dx
      
= 2k 0  dx/(t2+k2) (t2+1)  dx = /(k+1)
 
Now integrating from 1 to k and using fundamental theorem of calculus for a particular value of k = m
 
I(m) - I(1) =  1m I'(k) dk =  1m  /(k+1) dk = ln(m+1) - ln2
 
Now I(1) = 0/2  log(sin2x+cos2x) dx = 0 
 
Hence expressing the function in terms of k,
 
I(k) = ln(k+1) - ln2
 
PS: I am not quite sure what the forum expert was trying to do with that monograph on definite integrals. I hope he wasnt trying to ridicule kislay for a really instructive question.

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