help for this difficult integral?

Question

Integral [ square root ( sec^3(θ) * tanθ) ]

sec= 1/cos

Answer 1

Your integrand is (1/cos^3(x))*(sin(x)/cos(x))dx, or
(cos^(-4)(x))*sin(x)dx. Use substitution: let u = cos (x).

Answer 2

I fanally found your cure :p

lets say your integral is A

A = sqrt(sin^1/2(θ))dθ/cos^2(θ)

dθ/cos^2(θ)=dv, sin^1/2(θ)=u

tan(θ)=v, cos(θ)dθ/2sin^1/2(θ)=du

A = tan(θ)sin^1/2(θ) - 1/2 int sin^1/2(θ) dθ

i couldn't manage to solve the sin^1/2(θ) part but instead found smt better:

http://integrals.wolfram.com/index.jsp

good luck...

Answer 3

[ square root ( sec^3(θ) * tanθ) ]=sqrt[sec^2(θ)*sec(θ) * tanθ)]
=sec(θ)sqrt[sec(θ) * tan(θ)]
=sec(θ)sqrt[{1/cos(θ)} * {sin(θ)/cos(θ)}
=sec(θ)sqrt[sin(θ)/cos^2(θ)]
=sec^2(θ)sqrt{sin(θ)] now
Integral [ square root ( sec^3(θ) * tanθ) ]=
Integral [sec^2(θ)sqrt{sin(θ)]
Unfortunately sqrt{sin(θ)} is not an elementary function hence it's antiderivative does not exist.