Question

Evaluate the integral using the indicated trigonometric substitution. \int \frac{dx}{x^2\sqrt{x^2 - 9}}, \quad x = 3 \sec \theta \frac{\sqrt{x^2 - 9}}{x} + C \frac{1}{9} \frac{\sqrt{x^2 - 9}}{x} + C 9 \frac{\sqrt{x^2 - 9}}{x} + C \frac{1}{9} \frac{x}{\sqrt{x^2 - 9}} + C \frac{9x}{\sqrt{x^2 - 9}} + C

          Evaluate the integral using the indicated trigonometric substitution.
\int \frac{dx}{x^2\sqrt{x^2 - 9}}, \quad x = 3 \sec \theta
\frac{\sqrt{x^2 - 9}}{x} + C
\frac{1}{9} \frac{\sqrt{x^2 - 9}}{x} + C
9 \frac{\sqrt{x^2 - 9}}{x} + C
\frac{1}{9} \frac{x}{\sqrt{x^2 - 9}} + C
\frac{9x}{\sqrt{x^2 - 9}} + C

Show more…

Evaluate the integral using the indicated trigonometric substitution.
∫(dx)/(x^2√(x^2 - 9)), x = 3 secθ(√(x^2 - 9))/(x) + C
(1)/(9) (√(x^2 - 9))/(x) + C
9 (√(x^2 - 9))/(x) + C
(1)/(9) (x)/(√(x^2 - 9)) + C
(9x)/(√(x^2 - 9)) + C

Added by Ricardo O.

Calculus: Early Transcendentals

James Stewart 8th Edition

Thanks for your feedback!

Evaluate the integral using the indicated trigonometric substitution. int (dx)/(x^(2)sqrt(x^(2)-9)),x=3sec heta (sqrt(x^(2)-9))/(x)+C (1)/(9)(sqrt(x^(2)-9))/(x)+C 9(sqrt(x^(2)-9))/(x)+C (1)/(9)(x)/(sqrt(x^(2)-9))+C 9(x)/(sqrt(x^(2)-9))+C Evaluate the integral using the indicated trigonometric substitution. dx x=3sec x2v29 x29 +C O x 1x2=9 +C 19 O x -9 o+ 1 x b+ O x 9 V2 O 0+

Play audio

Feedback

Powered by NumerAI

Robert Daugherty and 61 other subject Calculus 1 / AB educators are ready to help you.

Ask a new question

Labs

Want to see this concept in action?

NEW

Explore this concept interactively to see how it behaves as you change inputs.

View Labs

Key Concepts

Recommended Videos

Recommended Textbooks

Transcript

00:01 Section 8 .4, problem number 8.

00:04 So we're dealing with integrals that require a trig substitution.

00:08 So integral 0 to 3 halves, dx 9 minus x squared to the 3 halves power.

00:15 So the substitution to make in this case is to let x equal 3 sine of theta.

00:25 Then it follows that dx would be equal to 3 cosine of theta.

00:33 And then let's work on 9 minus x squared to the three halves.

00:40 That's going to be 9.

00:43 And then when you square x, it's going to be, what, 9 sine squared.

00:47 So this is going to end up being 9.

00:50 1 minus the sine squared of theta to the 3 halves.

00:57 So this is 9 to the 3 halves, cosine squared of theta to the 3 halves.

01:06 So this ends up being, what, 27 cosine to the cubed power of theta.

01:16 So when, now let's just change the limits of integration.

01:20 When x is equal to zero, we have to look right here.

01:25 That means that the sine of zero, that means that zero is equal to three sine of theta, which tells me that theta is equal to zero.

01:36 So when x is equal to three halves, it tells me three halves is equal to three.

01:46 Sin theta, so sine of theta is one -half.

01:50 So theta is equal to pi over six...

Ace is your personal tutor. It breaks down any question with clear steps so you can learn.

Start Using Ace

Continue solving this problem

Create a free account to:

View full step-by-step solution
Ask follow-up questions with Ace AI
Save progress and study later