What is the solution to integrating \int\cos^{3/2}x \ dx?

[X-43D]

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]

what are you trying to find out?

 

[dextercioby]

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]

Thanks for the solution. I see that the 2nd is a generalized hypergeometric function. Is $$\int ( cos x )^{3/2} dx$$ also hypergeometric?

 

[dextercioby]

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.