Calculate the following integral:
∫ (square root(9x^2 - 16)) / x^2
I found:
∫ (square root((3x)^2 - (4)^2)) / x^2
x = a sec0
3x = 4 sec0
x = 4/3 sec0
dx = 4/3 sec0tan0 d0
4/3 ∫ [(square root(4sec0)^2 - 16)(sec0tan0 d0)] / x^2
I'm lost from here and don't know how to simplify.
∫√(9x^2 - 16) / x^2 dx
let 3x = 4secθ
so,
9x^2 = 16sec^2 θ
9x^2 - 16 = 16sec^2 θ - 16 = 16tan^2 θ
√(9x^2 - 16) = 4tanθ
3 dx = 4secθtanθ dθ
x^2 = 16/9 sec^2 θ
∫√(9x^2 - 16) / x^2 dx
= ∫4tanθ / (16/9 sec^2 θ) * (4/3 secθtanθ) dθ
= ∫3tan^2 θ secθ dθ
Use integration by parts on that. Double check my math here.
Thank you Steve
To simplify the integral, we can use a trigonometric substitution. Let's substitute x = (4/3)secθ and dx = (4/3)secθtanθ dθ.
After making the substitution, the integral becomes:
4/3 ∫ [(√(4secθ)^2 - 16)(secθtanθ dθ)] / [(4/3)secθ]^2
Now, simplify the integrand:
4/3 ∫ [(√(16sec^2θ - 16))(secθtanθ dθ)] / [(4/3)^2sec^2θ]
4/3 ∫ [√(16(sec^2θ - 1))(secθtanθ dθ)] / (16/9)sec^2θ
4/3 * (9/16) ∫ [√(sec^2θ - 1))(secθtanθ dθ)] / sec^2θ
(3/4) ∫ [(√(sec^2θ - 1))(secθtanθ dθ)] / secθ
Now, simplify further:
(3/4) ∫ [√(sec^2θ - 1))(tanθ dθ)]
Applying the identity sec^2θ - 1 = tan^2θ, we get:
(3/4) ∫ [√(tan^2θ))(tanθ dθ)]
(3/4) ∫ [(tanθ)(tanθ dθ)]
(3/4) ∫ [tan^2θ dθ]
The integral of tan^2θ can be evaluated using the power reduction formula for integrals:
∫ tan^2θ dθ = tanθ - θ + C
Therefore, the final result is:
(3/4)(tanθ - θ) + C
Now, substituting back x = (4/3)secθ:
The final answer is:
(3/4)(tanθ - θ) + C, where θ = sec^(-1)(3x/4) and C is the constant of integration.