- #176
royjsky
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zoki85 said:
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.
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.
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.
- #177
royjsky
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I think you need multivariable to do that.DyslexicHobo said:
- #178
Corbeau
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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 ) )
1/ ( x ( ln (x+1) -1 ) )
- #179
suprabhe
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x^2arctan(5x)
- #180
- 14,132
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Last edited by a moderator:
- #181
MAGNIBORO
- 106
- 26
$$\int_{0}^{\pi}\frac{cos(nx)-cos(na)}{cos(x)-cos(a)} dx$$
and
$$\int_{0}^{\infty }\frac{x}{e^{x}-1} dx$$
and
$$\int_{0}^{\infty }\frac{x}{e^{x}-1} dx$$
Last edited:
- #182
Suraj Pal
Try this.
- #183
Ritam Dutta
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it should come from common sense i think. seems the analytical method is going to be just WOOWzoki85 said:
- #184
ComplexVar89
Gold Member
- 21
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I was just thinking this thread is quite reminiscent of that book (without Nahin's witty commentary, of course.)Demystifier said:
- #185
- #186
quarantine2020
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Ferhat said:
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
Ssnow
Gold Member
- 570
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Try:
[tex] \int_{0}^{1} e^{-x^{x}} dx [/tex]
Ssnow
[tex] \int_{0}^{1} e^{-x^{x}} dx [/tex]
Ssnow
- #188
akil
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try the integral of sin(lnx) by using eulers formula
- #189
Aritro Biswas
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The answer isyip said:
[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
Ossi
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Is It -sqrt(pi/2) took me a bit to calculate. Its doable, If one knows the trickszoki85 said:
- #191
mathhabibi
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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
##\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
Vanadium 50
Staff Emeritus
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As my first calculus teacher said, "there is a difference between a hard problem and a long problem."
- #193
mathhabibi
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Is this directed toward my integral? If it is, I could take it down from this thread.Vanadium 50 said:
- #194
mathhabibi
- 48
- 13
Actually, I can't delete that post.mathhabibi said:
- #195
mathhabibi
- 48
- 13
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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