Question

Solve each of the integrals . Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible. $$ \int x \sin ^{-1} x d x $$

Name: Problem 74, Chapter 5: Calculus: Early Transcendentals
Uploaded: 2022-02-21T02:47:44-08:00
Duration: 3 min 47 s
Channel: Linda Hand
Description: Solve each of the integrals . Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible. ∫x sin^-1 x d x

   Solve each of the integrals . Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.
$$
\int x \sin ^{-1} x d x
$$

Calculus: Early Transcendentals

Michael Sullivan,… 1st Edition

Chapter 5, Problem 74

Instant Answer

Solved by Expert Linda Hand

02/21/2022

Step 1

We choose $u = \sin^{-1}x$ and $dv = x dx$. Then $du = \frac{1}{\sqrt{1-x^2}} dx$ and $v = \frac{x^2}{2}$. Show more…

Show all steps

Thanks for your feedback!

Solve each of the integrals . Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible. $$ \int x \sin ^{-1} x d x $$

Play audio

Feedback

Linda Hand and 65 other subject Calculus 2 / BC 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

Properties of Inverse Trigonometric Functions

Inverse trigonometric functions, such as arcsin(x), have well-defined derivatives that are crucial when applying integration techniques like integration by parts. Understanding these derivatives, for example, d/dx(arcsin(x)) = 1/?(1 ? x²), aids in selecting the appropriate components for u and dv and simplifies the integration process.

Integration by Parts

Integration by parts is a technique used to evaluate integrals of products of functions. It is based on the product rule for differentiation and is expressed as ? u dv = uv ? ? v du. This method is particularly useful when one part of the integrand simplifies upon differentiation, allowing the integral to be reduced to a simpler form.

Please give Ace some feedback