Challenging Integrals in Calculus 1-2: Expand Your Problem-Solving Skills!

In summary: The integral can be evaluated using a change of coordinates to polar coordinates and the squeeze theorem.
  • #176
zoki85 said:
Hard ,but famous and bautiful :

[tex]\int_{0}^{\infty}sin(x^2)dx[/tex]

I believe this uses multivariable calculus since I recall you must first prove the definite integral of the e^-x^2 which requires double integrals and a change of variables.
 
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  • #177
DyslexicHobo said:
I'm just out of Calc 1, so I'm not sure if I even have the knowledge to solve this... but here's where I am now. I can't tell if I complicated it even more, or if I'm closer to getting the solution.

[tex]\int (r-1)^{1/5}ln(r) - r^{-1} (r-1)^{1/5} dr[/tex]

I used u-substitution (well, r-substitution), where [tex]r = x^5 + 1[/tex]. After the substitution I used integration by parts, and now I'm unsure if that was even the right path. Please let me know!

And please tell us how to do [tex]\int_{0}^{\infty}sin(x^2)dx[/tex]
I think you need multivariable to do that.
 
  • #178
Here is a hard nut to crack. Been trying to solve it for a while, can't confirm it's actually solvable.

1/ ( x ( ln (x+1) -1 ) )
 
  • #179
x^2arctan(5x)
 
  • #181
$$\int_{0}^{\pi}\frac{cos(nx)-cos(na)}{cos(x)-cos(a)} dx$$

and

$$\int_{0}^{\infty }\frac{x}{e^{x}-1} dx$$
 
Last edited:
  • #182
Try this.
1507919193165-1998323113.jpg
 
  • #183
zoki85 said:
Hard ,but famous and bautiful :

[tex]\int_{0}^{\infty}sin(x^2)dx[/tex]
it should come from common sense i think. seems the analytical method is going to be just WOOW
 
  • #185
yip said:
Try [tex] \int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}} [/tex]
(forgot to put the integral sign in, it is now fixed)

This is my answer, tell me if i did something wrong :).
 

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  • #186
Ferhat said:
This is my answer, tell me if i did something wrong :).

1589576667909.png

this is a “simpler” result I got from unraveling your solution & re-packaging it. I want to figure out how to get it in this form in a more natural way. As of now I’m stuck. Right now I’m working with a Pythagorean triangle
Adjacent = 1-x^2
Opposite = x*sqrt(2)
Hypotenuse = sqrt(1+x^4)

and the solution is 1/sqrt(2)*ln(sec(angle)+tan(angle)) + C

I see some kind of pattern here but it’s a little opaque. Any way to clear this up & produce a really elegant solution?
 
  • #187
Try:

[tex] \int_{0}^{1} e^{-x^{x}} dx [/tex]

Ssnow
 
  • #188
try the integral of sin(lnx) by using eulers formula
 
  • #189
yip said:
Try [tex] \int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}} [/tex]
(forgot to put the integral sign in, it is now fixed)
The answer is
[tex]
\frac{1}{2\sqrt{2}}\ln \left| { \frac{\sqrt{2}+{\sqrt{x^2+\frac{1}{x^2}}}} {\sqrt{2}-{\sqrt{x^2+\frac{1}{x^2}}}}} \right|
[/tex]
 
  • #190
zoki85 said:
Hard ,but famous and bautiful :

[tex]\int_{0}^{\infty}sin(x^2)dx[/tex]
Is It -sqrt(pi/2) took me a bit to calculate. Its doable, If one knows the tricks
 
  • #191
Try this one:
##\int_0^\infty\frac{\sin^2x}{x^2(x^2+1)}dx##
If you need an explanation, let me know. But I want to give you guys some time to find out how to do it
 
  • #192
As my first calculus teacher said, "there is a difference between a hard problem and a long problem."
 
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  • #193
Vanadium 50 said:
As my first calculus teacher said, "there is a difference between a hard problem and a long problem."
Is this directed toward my integral? If it is, I could take it down from this thread.
 
  • #194
mathhabibi said:
Is this directed toward my integral? If it is, I could take it down from this thread.
Actually, I can't delete that post.
 
  • #195
Here's another integral that I find interesting $$\int_0^\infty\frac{\sin x}{\sinh x}dx$$This one has an answer in terms of hyperbolic cotangent.
 

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