Trigonometric Substitution Integration

Question

4.) ∫((9-25x^2)^(1/2)*dx)/(x^2)

Answer 1

Here is my solution and answer to your question in standard format so you can easily understand. Go to this link:

https://drive.google.com/file/d/1jXdP57IHkY0iMrmMRkMAISShrWrWjNmL/view?usp=sharing

Let me know if you have any question.

Answer 2

Make the substitution x=(3/5) sin θ. This should get you an integrand which will integrate as an arcsin.

Answer 3

My solution and answer is found in this link:

https://drive.google.com/file/d/1jXdP57IHkY0iMrmMRkMAISShrWrWjNmL/view?usp=sharing

I did it this way because there several mathematical symbols that I can't use directly in Wyzant Ask The Expert. Let me know if you have any question.

Answer 4

Let x = (3/5)secθ. Then, dx = (3/5)secθtanθdθ

So, the integral becomes ∫[(3/5)secθtanθdθ) / (9/25sec²θ√((9)(1 - sec²θ))

= (5/9)∫cosθdθ = (5/9)sinθ + C

Since x = (3/5)secθ, secθ = 5x/3. So, cosθ = 3/(5x) and sinθ = √(1 - cos²θ) = √(25x² - 9) / 5x

So, the final answer is √(25x² - 9) / (9x) + C

Answer 5

Substitution: x = 3/5sint; dx = 3/5costdt. So, our integral I = ∫(9 - 9sin²t)^1/23/5cost/9/25sin²tdt = 9/5·25/9∫cos²t/sin^{2tdt = 5}∫cot²tdt = 5∫(csc²t - 1)dt = - 5cott - 5t + C.

Now back to x: sint = 5x/3, t = sin^-1(5x/3); cott = (9 - 25x²)^1/2/5x.

Because our integral I = - (9 - 25x²)^1/2/x - 5sin^-1(5x/3) + C