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July 31, 2014

July 31, 2014

Posted by **Angel** on Tuesday, February 12, 2013 at 2:53pm.

Ingrate (x^3)/sqrt(x^2 - 9)

I know that x= 3sec(theta)

Sqrt(x^2-9)= 3tan(theta)

x^3 = 27Sec^3(theta)

Once I put that all together I'm not sure what to do next.

- Math (Cal) -
**Steve**, Tuesday, February 12, 2013 at 3:45pmso far, so good. But you forgot about dx:

dx = 3secθtanθ dθ

So, now you have

∫27sec^3(θ) / 3tanθ * 3secθtanθ

= 27∫sec^4(θ) dθ

= 27∫(tan^2(θ)+1) sec^2(θ) dθ

Now if you let u = tanθ, you have

= 27∫(u^2+1) du

= 27(u^3/3 + u)

= 27(tan^3(θ)/3 + tanθ)

= 9tanθ (tan^2(θ) + 3)

Now, the original substitutions,

tanθ = 1/3 √(x^2-9), so we wind up with

3√(x^2-9) (1/9 (x^2-9) + 3)

= 1/3 √(x^2-9) (x^2-9+27)

= 1/3 √(x^2-9) (x^2+18)

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