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Was trying to see if I can get the reduction formulas for int(cos(x)^n,x) in maple. But it seems no assumption used can make Maple give any result for this.  Mathematica gives a result using Hypergeometric2F1 (even with no assumption on n, which I am not sure about now), but was wondering why maple can't do this one:


int( (cos(x))^n,x) assuming n::integer;

int( (cos(x))^n,x) assuming n::posint;

In Mathematica, I get:

I am newbie in Maple, so may be I am missing some command or doing something wrong.

ps. I was trying to obtain

But this is lost case now. I just need to find out first why int(cos(x)^n,x) does not evaluate to anything in Maple.

fyi, the Hypergeometric result for $\int cos^n(x) \,dx$ can be seen in this reference (half way down the page):

ps. can't one enter Latex in this forum like at stack exchange?



Hi there

How can I enter an arbitrary partition on an interval for calculating a Riemann sum by Maple 13?please write the concerning command.



Dear all,

I woul line integrate e gradient moltiply to a vector. I will try to explain better.

I have define this function:


and I have a vector a


I would like to applu a dot product between the gradient of phi an the vector and integrate the results. 

I have already try in many way but without succeed. Someone could please help me?


I have updated  Maple from 18.01 to 18.02, but there is something strange happened to me. I can not use int anymore. Here is my codes:

Error, (in int) wrong number (or type) of arguments: invalid option value passed to indefinite integration: {}

print(`output redirected...`); # input placeholder
    Maple 18.00, X86 64 LINUX, Feb 10 2014, Build ID 922027


Here is the screenshot:


I have a question: Why gives Maple no result for:

expr2 := sin(k*(t-tau))*sin(k*tau);
Int(simplify(expr2, symbolic), tau = 0 .. t);

It is possible with Wolfram Alpha:*%28t-tau%29%29*sin%28sqrt%28k%2Fm%29*tau%29%2C+tau+%3D+0+..+t%29



int(int(y, 0 <= y, x^2+y^2+z^2 <= 1));

Error, (in int) integration range or variable must be specified in the second argument, got 0 <= y

int(int(y, y = 0, x^2+y^2+z^2 = 1));

Error, (in int) integration range or variable must be specified in the second argument, got y = 0

Hi all

kx,ky is the wavenumber, how can I get the 4 cases of piecewise function according to kx=0,kx≠0 and ky=0,ky≠0. Thanks

J := `assuming`([4*(int(int(JJ*exp(-I*(kx*x+ky*y))*sin(2*l*pi*x/a)*sin(2*k*pi*y/b), x = 0 .. a, AllSolutions), y = 0 .. b, AllSolutions))/(a*b)], [k::posint, l::posint, a > 0, b > 0, JJ > 0])

Hi all

Assume that we have following vectors:

> V1 := [1/9, -5/9, 7/9, 1/9, -5/9, 7/9, 1/9, -5/9, 7/9];

>  V2:=t*V1;

and we want to compute the integral of V2, namely:

>  Int(seq(V2[i],i=1..9),t=0..1);

            [       5 t  7 t         5 t  7 t         5 t  7 t]
      V2 := [t/9, - ---, ---, t/9, - ---, ---, t/9, - ---, ---]
            [        9    9           9    9           9    9 ]

Error, (in print/Int) invalid input: IntegrationTools:-GetOptions expects its 1st argument, v, to be of type Integral, but received Int(1/9*t,-5/9*t,7/9*t,1/9*t,-5/9*t,7/9*t,1/9*t,-5/9*t,7/9*t,t = 0 .. 1)

 how we integrate from V2? why answer is wrong in my code?

thanks a lot

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department

I have a great problem with this integral and Maple gives two answers completely different:


int(x^-5/3*cos((x-1)*h), x = 0..infinity)

so I get two different results :


-(27/8)*h^2+3/2+(27/8)*h^(7/6)*LommelS2(11/6, 1/2, h)


or this:


-(27/8)*h^2+3/2+(27/8)*h^(7/6)*LommelS1(11/6, 1/2, h)

In the first integral A get Lommels2 and If I get the Integral by using Taylor of cos((x-1)*h) and after that I resum I get Lommels1.


Thank you.



my int is in below.

my int has hyperbolic and trigonal parts thus when they are multiplied , Maple is not able to integrate!


Now, How can I solve this integral?

thank you for help

my int is:




eq1 := m*(diff(diff(w(x, t), t), t))+diff(diff(EIy*(diff(diff(w(x, t), x), x))+(EIz-EIy)*(diff(diff(w(x, t), x), x))*(theta(x)+phi(x, t))^2, x), x)-Pz = 0

l := 16; m := .75; EIy := 0.2e5; EIz := 0.400e7; GJ := 0.1e5; mj := .1; Pz := 5000; Mz := 0; theta := proc (x) options operator, arrow; 0 end proc

w := proc (x, t) options operator, arrow; q[1](t)*(cosh(1.8751*x/l)-cos(1.8751*x/l)+(-1)*.734096*(sinh(1.8751*x/l)-sin(1.8751*x/l)))+q[2](t)*(cosh(4.6941*x/l)-cos(4.6941*x/l)+(-1)*1.01847*(sinh(4.6941*x/l)-sin(4.6941*x/l)))+q[3](t)*(cosh(7.8548*x/l)-cos(7.8548*x/l)+(-1)*.999224*(sinh(7.8548*x/l)-sin(7.8548*x/l))) end proc

phi := proc (x, t) options operator, arrow; q[4](t)*sqrt(2)*sin(1.5708*x/l)+q[5](t)*sqrt(2)*sin(4.7124*x/l)+q[6](t)*sqrt(2)*sin(7.8540*x/l) end proc


f[1] := int(lhs(eq1)*(cosh(1.8751*x/l)-cos(1.8751*x/l)-.734096*(sinh(1.8751*x/l)-sin(1.8751*x/l))), x = 0 .. l)


Hi all,

As we know that the indefinite integral in Maple is defined up to a piecewise constant.

For example,


Due to the indeterminate range of variable n, int returns the piecewise function.

But in the other similar case, we get a generic solution rather than a piecewise function,


Why not to returns the result like this,


Thanks for any help.

I have an expression like this:

Since it is linear I want Maple to rewrite it into this:

(with the benefit that Maple then can solve it at least up to a point). i have tried to conceive a rule to do that but got stuck relatively quickly. Does anybody have a way to do this (in some genrality)?


Mac Dude.


limit(t*(int(exp(-t*tan(x)), x = 0 .. (1/2)*Pi)), t = infinity)?

Is it possible to find it in Maple? The command

MultiSeries:-limit(t*(int(exp(-t*tan(x)), x = 0 .. (1/2)*Pi)), t = infinity);









I am unable to plot DP3 vs Q, not only it take very long time but didn't give any output.

Please have a look.




firstly apologies in advance for stuff in this question such as "triangle symbol",  my computer is pretty old. 

ok so i was confused a bit here, what i'm trying to do is write a maple procedure that computes Af for a given f contained in V . except we only need to correct the bug in the script below. This script demonstrates such a procedure in the case that omega is a square. The domain is given here as the negative set of a function F contained in V .  I have left in notes where/what i think we need to do but i dunno how to...

N:=10 ; # Global Var
Average := proc(F, f0) local f, i, j;
f := f0; # !!!!!!!!!!!!!! something is bad here...
for i to N do for j to N do
if F(i, j) < 0 then
f[i, j] := (f0[i - 1, j] + f0[i + 1, j] + f0[i, j + 1] + f0[i, j - 1])/4 ;
end if;
end do;end do;
return f;
end proc;
f0:=Matrix(N,F); # just to have something to test the procedure
Average(F,f0); # does not return the expected average, modifies f0


the necessary information we were given to produce this so far was..

Let N be a positive integer and [N] = {i contained in N | 1<= i <=N }  Let "Omega" C {(i,j) contained in [N] x [N] | 2<=i,j<=N-1} be a subset. Let V = R^([N]x[N]) be the vector space of real valued functions [N]x[N] -> R
and A, "triangle symbol":V->V (average) and "triangle symbole" (Laplacian) be the linear maps such that
[Af](i; j) = f(i; j)      if (i; j) not contained in "Omega"   OR

                             [f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 if (i,j) is contained in "Omega"

["traingle symbol"f](i,j) =  0 if (i,j) isnt contained in "Omega"   OR

                            ( f(i,j) - [f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 )    if (i,j) is contained in "Omega"

 Please and thank you for any help in advance <3


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