- #36
rocomath
- 1,755
- 1
is this a Calc 1 or 2 problem?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)
is this a Calc 1 or 2 problem?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)
i thought so, because i didn't have a prob like that in Calc 1 ... I'm dying to know though, lol :rofl:HallsofIvy said:I would consider it Calc II.
Gib Z said:[tex]\int^{1}_0 \frac{\log_e (1+x)}{x} dx [/tex]. Quite an interesting one that someone gave to me. Nice Solution :)
Kummer said:I much more interesting integral is,
[tex] \int_0^{\infty} \cos x^k dx = \cos \frac{\pi}{2k} \cdot \Gamma [(k+1)/k][/tex]
And,
[tex]\int_0^{\infty} \sin x^k dx = \sin \frac{\pi}{2k} \cdot \Gamma [(k+1)/k][/tex]
-Wolfgang
Klaus_Hoffmann said:[tex] \sqrt tan(x) [/tex] can't be reduced to Rational functions or known functions
lurflurf said:[tex]\int \sqrt{\tan(x)}dx=\int\frac{\sec^2(x)}{1+\tan^2(x)}\sqrt{\tan(x)}dx=\int\frac{2\tan(x)}{1+\tan^2(x)}d\sqrt{\tan(x)}[/tex]
so it hinges on the always fun
[tex]\int\frac{x^2}{1+x^4}dx[/tex]
yip said:For that integral, i think u have to use the partial fraction theorem that involves derivatives, as there is trigonometric terms in the primitive of that function. Here is how I did the second integral, since u asked, the trick is to apply euler's theorem:
[tex] e^{-ix^{2}}=cosx^{2}-isinx^{2}[/tex]
[tex]\int_{0}^{\infty}e^{-ix^{2}}dx=\int_{0}^{\infty}cosx^{2}dx-i\int_{0}^{\infty}sinx^{2}dx[/tex]
It is known that [tex]\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}[/tex]
[tex]\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{\sqrt{\pi}}{2\sqrt{i}}[/tex]
[tex]\frac{1}{\sqrt{i}}=\frac{1-i}{\sqrt{2}}[/tex]
[tex]\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}(1-i)[/tex]
[tex]\int_{0}^{\infty}sinx^{2}dx=I\int_{0}^{\infty}e^{-ix^{2}}dx=\frac{1}{2}\sqrt{\frac{\pi}{2}}[/tex]
jostpuur said:However, the Leibniz's rule of alternating series can be used to show that integrals
[tex]
\int\limits_{0}^{\infty}\cos(x^2)dx
[/tex]
and
[tex]
\int\limits_{0}^{\infty}\sin(x^2)dx
[/tex]
converge towards something, and thus the integral
[tex]
\int\limits_{-\infty}^{\infty} e^{ix^2}dx
[/tex]
also exists. If we know that the integral is a continuous function of the coefficent in the exponent, then this is all done, but I never bothered actually try to figure out how the continuity could be proven. If it can be done nicely, feel free to mention how.
jostpuur said:Or actually I'm not so sure about this all yet. The derivation that uses gamma functions seems to assume right in the beginning that integral
[tex]
\int_{0}^{\infty} e^{ix^k} dx
[/tex]
converges. It would be nice to show it some Leibniz's test, for convergence, before continuing with variable changes, that are used for solution of the value of the integral.
Isn't it just 0?JohnDuck said:I'm stumped but intrigued.
camilus said:[tex]I =\int {sinxcosx \over sin^4x+cos^4x}dx[/tex]