Trig. Substitution: integral [x^3/(sqrt.(x^2-4))]dx

[Guest]

I got to a certain point and then got stuck. Here is the problem:

. . .integral [x^3/(sqrt.(x^2-4))]dx

My work:

Let u = x, u = asec(A), a = 2, x = 2sec(A), sec(A) = x/2

In a right triangle, let the hypotenuse have length x, the adjacent side have length 2, and the opposite side have length sqrt(x^2 - 4). Let:

. . .idx = 2sec(A) tan(A) dA

Now the integral becomes:

. . .iintegral of [(2sec(A))^3(2sec(A) tan(A))]/2 tan(A) dA

This reduces to:

. . .iintegral of 8(secA)^4dA

. . .iintegral of 8 (sec^2A)^2dA

. . .iintegral of 8(1/2(tan2A+1)^2)dA

. . .iintegral of 4(tan2A)^2+2tan2A+1)dA

. . .iintegral of 2((sec4A-1)+2tan2A+1)dA

. . .iintegral of (2sec4A-2+4tan2A+2)dA

This is where I get stuck. Is my work so far correct? And if so, where do I go from here? Thank you!

 

[galactus]

\(\displaystyle \L\\\int{sec^{n}(x)}dx=\frac{sec^{n-2}(x)tan(x)}{n-1}+\frac{n-2}{n-1}\int{sec^{n-2}(x)}dx\)