Help in integration

  • #1
fantomas2005
4
0
can someone integrate that problem

x^2 cos (x^5) dx

i tried integration by parts but i didn't get anything
 

Answers and Replies

  • #2
rootX
465
4
There might be something wrong with the question if you are in an introductory course

some mess:

f =

x^2*cos(x^5)


>> int(f,x)

ans =

1/10*2^(3/5)*pi^(1/2)*(5/3/pi^(1/2)/x^2*2^(2/5)
*sin(x^5)+5/3/pi^(1/2)/x^2*2^(2/5)*(cos(x^5)
*x^5-sin(x^5))-2/3/pi^(1/2)*x^8*2^(2/5)/(x^5)^(11/10)
*sin(x^5)*LommelS1(1/10,3/2,x^5)-5/3/pi^(1/2)*x^8
*2^(2/5)/(x^5)^(21/10)*(cos(x^5)*x^5-sin(x^5))
*LommelS1(11/10,1/2,x^5))


>> f = x^2*(cos(x))^5

f =

x^2*cos(x)^5


>> int(f,x)

ans =

x^2*(1/5*cos(x)^4*sin(x)+4/15*cos(x)^2*
sin(x)+8/15*sin(x))+2/25*x*cos(x)^5-2/125*
cos(x)^4*sin(x)-272/3375*cos(x)^2*sin(x)-4144/3375*
sin(x)+8/45*x*cos(x)^3+16/15*x*cos(x)
 
  • #3
fantomas2005
4
0
no there is nothing wrong with the problem and iam in what called in our country differentiation and integration 2 like calculus 3
and the original problem is from my exam {{ and i didn't solve it but I'm curious }}



given____ { int from 0 to 4 }{int from y^1/2 to 2 } cos(x^5) dx dy ____ reverse the order of integration and evaluate the resulting intgerals .


after reversing , it will be ______ x^2 cos (x^5) dx _______ and the inequalities dosent matter
 
  • #4
Vid
402
0
You reversed the order of integration. You intergrate with wrt y first, dydx.
 
Last edited:
  • #5
fantomas2005
4
0
so what is the solution of the integration pls
 
  • #6
Kurdt
Staff Emeritus
Science Advisor
Gold Member
4,846
6
Are you sure it was [itex] \cos(x^5)[/itex]? You really couldn't integrate that on a test. The method for reversing the order of integration is fairly simple. You first need to sketch the area of integration on a graph and then find the limits for integrating over y with x constant and then the limits for integrating over x with y constant.
 
  • #7
fantomas2005
4
0
i'm absolutely sure that it's int of x^2 cos(x^5) dx

dose anyone have any idea ??
 
  • #8
dirk_mec1
761
13
What was the original question? We've already proved that the integral you post is too difficult!
 

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