Trigonometric substitution

Question

My daughter is having problems with this question. Any help is appreciated.

a. Calculate the integral ((SR9x^2-4)/x^3)dx by using trigonometric substitution.

b. Determine the Integrand obtained after substituting x and simplifying. Then evaluate the integral. Answer should be in terms of theta.

c. Determine the value of the original integral in terms of x

integral ((SR9x^2-4)/x^3)dx = ?

Answer 1

Answer 2

Answer 3

∫[√(9x²-4)/x³]dx Letting u = 3x,

the integral becomes 9∫[√(u²-4)/u³]du which is a little bit easier to work with.

A. Using Trigonometric Substitution

u = 2 sec θ, du = 2 sec θ tan θ dθ, and √(u²-4) = 2 tan θ

With this trig substitution

9∫[√(u²-4)/u³]du becomes (9/2)∫(tan²θ/sec²θ)dθ = (9/2)∫sin²θ dθ

This sin²θ integral can be solved using integration by parts, but there's a quicker, less laborious way to get the same result

Since cos 2θ = 1 – 2sin²θ, we can substitute sin²θ = ½(1 – cos 2θ)

and our integral becomes

(9/2)∫[½(1 – cos 2θ)] dθ = (9/4)∫(1 – cos 2θ) dθ =

(9/4)θ – (9/8)sin 2θ

(I'm going to omit the constant of integration until the end)

Next we substitute sin 2θ = 2(sin θ)(cos θ), so

(9/4)θ – (9/8)sin 2θ becomes (9/4)θ – (9/4)(sin θ)(cos θ)

B. Evaluation of the Integral in Terms of θ

(9/2)∫(tan²θ/sec²θ)dθ = (9/2)∫sin²θ dθ = (9/4)θ – (9/4)(sin θ)(cos θ) + C

C. Back to x Again

Now we substitute back to put everything in terms of x.

Since u = 3x = 2 sec θ this means sec θ = 3x/2 and

θ = sec^–1(3x/2)

cos θ =1/sec θ = 2/3x

sin θ = √(1 – cos²θ) = √(1 – cos²θ) = √(9x²-4)/3x

and (sin θ)(cos θ) = 2√(9x²-4)/9x²

It can be helpful here to draw a right triangle with side lengths marked as

hypotenuse = 3x

adjacent side (to θ) = 2

opposite side (to θ) = √(9x²-4)

The final result is

∫[√(9x²-4)/x³]dx = (9/4)sec^–1(3x/2) – √(9x²-4)/2x² + C

I hope this helps.