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.

Mark C.

asked • 06/30/20**4.) ∫((9-25x^2)^(1/2)*dx)/(x^2)**

Follow
•
1

Add comment

More

Report

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.

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.

Mark M. answered • 06/30/20

Tutor

4.9
(882)
Retired math prof. Calc 1, 2 and AP Calculus tutoring experience.

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

So, the integral becomes ∫[(3/5)secθtanθdθ) / (9/25sec^{2}θ√((9)(1 - sec^{2}θ))

= (5/9)∫cosθdθ = (5/9)sinθ + C

Since x = (3/5)secθ, secθ = 5x/3. So, cosθ = 3/(5x) and sinθ = √(1 - cos^{2}θ) = √(25x^{2} - 9) / 5x

So, the final answer is √(25x^{2} - 9) / (9x) + C

Yefim S. answered • 06/30/20

Tutor

5
(20)
Math Tutor with Experience

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

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

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

Ask a question for free

Get a free answer to a quick problem.

Most questions answered within 4 hours.

Find an Online Tutor Now

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.