Find int(1)/(sqrt(-x^(2)+12 x-32))dx. Choose 1 answer: (A) arcsin((x-6)/(2))+C (B) (1)/(12)arctan((x-6)/(2))+C (C) (1)/(12)arcsin((x-6)/(2))+C (D) arctan((x-6)/(2))+C

Substitute u: Now, let's substitute u = (x - 6)/2, which means du = 1/2 dx or dx = 2 du. The integral becomes int 1/sqrt(4 - u^2) * 2 du.Simplify integral: Simplify the integral to get 2 * int 1/sqrt(4 - u^2) du. Recognize that this is the integral form of arcsin(u/2), since the derivative of arcsin(u/2) is 1/sqrt(1 - (u/2)^2).Recognize integral form: Now, integrate to get 2 * arcsin(u/2) + C.Integrate to get: Substitute back for u to get 2 * arcsin((x - 6)/4) + C. But wait, there's a mistake here. The integral should be arcsin(u/2) without the 2 in front, since we already accounted for the dx = 2 du substitution.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Find
int(1)/(sqrt(-x^(2)+12 x-32))dx.
Choose 1 answer:
(A)
arcsin((x-6)/(2))+C
(B)
(1)/(12)arctan((x-6)/(2))+C
(C)
(1)/(12)arcsin((x-6)/(2))+C
(D)
arctan((x-6)/(2))+C

Find $\int \frac{1}{\sqrt{-x^{2}+12 x-32}} d x$ . $\newline$ Choose $1$ answer: $\newline$ (A) $\arcsin \left(\frac{x-6}{2}\right)+C$ $\newline$ (B) $\frac{1}{12} \arctan \left(\frac{x-6}{2}\right)+C$ $\newline$ (C) $\frac{1}{12} \arcsin \left(\frac{x-6}{2}\right)+C$ $\newline$ (D) $\arctan \left(\frac{x-6}{2}\right)+C$

View step-by-step help

Home

Math Problems

Calculus

Evaluate definite integrals using the chain rule

Full solution

Q. Find

\int \frac{1}{\sqrt{-x^{2}+12 x-32}} d x

\newline

Choose

1

answer:

\newline

(A)

\arcsin \left(\frac{x-6}{2}\right)+C

\newline

(B)

\frac{1}{12} \arctan \left(\frac{x-6}{2}\right)+C

\newline

(C)

\frac{1}{12} \arcsin \left(\frac{x-6}{2}\right)+C

\newline

(D)

\arctan \left(\frac{x-6}{2}\right)+C

Substitute $u$ : Now, let's substitute $u = \frac{x - 6}{2}$ , which means $du = \frac{1}{2} dx$ or $dx = 2 du$ . The integral becomes $\int \frac{1}{\sqrt{4 - u^2}} \cdot 2 du$ .
Simplify integral: Simplify the integral to get $2 \int \frac{1}{\sqrt{4 - u^2}} du$ . Recognize that this is the integral form of $\arcsin(\frac{u}{2})$ , since the derivative of $\arcsin(\frac{u}{2})$ is $\frac{1}{\sqrt{1 - (\frac{u}{2})^2}}$ .
Recognize integral form: Now, integrate to get $2 \cdot \arcsin(\frac{u}{2}) + C$ .
Integrate to get: Substitute back for $u$ to get $2 \cdot \arcsin\left(\frac{x - 6}{4}\right) + C$ . But wait, there's a mistake here. The integral should be $\arcsin\left(\frac{u}{2}\right)$ without the $2$ in front, since we already accounted for the $dx = 2 du$ substitution.