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X-43D
Mar25-05, 05:01 PM
The function of the type:

\int {(x^2 + 1)^{5/2}}x dx

This is simple to integrate but the trigonometric function:

\int 3/5{(\sec x)}^{5/3}x dx is already a problem.

The first gives:

\int {(x^2 + 1)^{5/2}}x dx = \int {u}^{5/2}1/2 du = 1/2 \int u^{5/2} du = 1/2 ({2u^{7/2}/7 + C) = 1/7{(x^2 + 1)}^{7/2} + C

p53ud0 dr34m5
Mar25-05, 06:50 PM
what are you trying to find out?

dextercioby
Mar26-05, 03:57 AM
Okay.There's no possible connection between the 2 integrals and the second is not an elliptical one.

There's the result for

\int x (\sec x)^{\frac{5}{3}} \ dx


Daniel.

X-43D
Mar26-05, 06:31 AM
Thanks for the solution. I see that the 2nd is a generalized hypergeometric function. Is \int ( cos x )^{3/2} dx also hypergeometric?

dextercioby
Mar26-05, 08:44 AM
Nope,that's elliptic.I think i've posted the solution in another thrread *looks for the solution*.Nope i confused it with another one.

There it is

\int \cos^{3/2}x \ dx

is equal to




Daniel.