Question

`\int\limits^a_b {(1-x^(2) )^(3/2) } \, dx`

Answer 1

First integrate the indefinite integral,

`\int(1-x^2)^(3/2)dx`

Let
`x=\sin(u)` which will make
`dx=\cos(u)du`.

Then

`(1-x^2)^(3/2)=(1-\sin^2(u))^(3/2)=\cos^3(u)` which makes
`u=\arcsin(x)` and our integral is reshaped,

`\int\cos^4(u)du`

Use reduction formula,

`\int\cos^m(u)du=(1)/(m)\sin(u)\cos^(m-1)(u)+(m-1)/(m)\int\cos^(m-2)(u)du`

to get,

`\int\cos^4(u)du=(1)/(4)\sin(u)\cos^3(u)+(3)/(4)\int\cos^2(u)du`

Notice that,

`\cos^2(u)=(1)/(2)(\cos(2u)+1)`

Then integrate the obtained sum,

`(1)/(4)\sin(u)\cos^3(u)+(3)/(8)\int\cos(2u)du+(3)/(8)\int1du`

Now introduce
`s=2u\implies ds=2du` and substitute and integrate to get,

`(3\sin(s))/(16)+(1)/(4)\sin(u)\cos^3(u)+(3)/(8)\int1du`

`(3\sin(s))/(16)+(3u)/(4)+(1)/(4)\sin(u)\cos^3(u)+C`

Substitute 2u back for s,

`(3u)/(8)+(1)/(4)\sin(u)\cos^3(u)+(3)/(8)\sin(u)\cos(u)+C`

Substitute
`\sin^(-1)` for u and simplify with
`\cos(\arcsin(x))=√(1-x^2)` to get the result,

`\boxed{(1)/(8)(x√(1-x^2)(5-2x^2)+3\arcsin(x))+C}`

Let
`F(x)=(1)/(8)(x√(1-x^2)(5-2x^2)+3\arcsin(x))+C`

Apply definite integral evaluation from b to a,
`F(x)\Big|_b^a`,

Hope this helps :)

Answer 2

Answer:
`\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{(3)/(2)} + 3a√(1 - a^2)}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{(3)/(2)} + 3b√(1 - b^2)}{8}`General Formulas and Concepts:

Pre-Calculus

• Trigonometric Identities

Calculus

Differentiation

• Derivatives
• Derivative Notation

Integration

• Integrals
• Definite/Indefinite Integrals
• Integration Constant C

Integration Rule [Reverse Power Rule]:
`\displaystyle \int {x^n} \, dx = (x^(n + 1))/(n + 1) + C`

Integration Rule [Fundamental Theorem of Calculus 1]:
`\displaystyle \int\limits^b_a {f(x)} \, dx = F(b) - F(a)`

U-Substitution

• Trigonometric Substitution

Reduction Formula:
`\displaystyle \int {cos^n(x)} \, dx = (n - 1)/(n)\int {cos^(n - 2)(x)} \, dx + (cos^(n - 1)(x)sin(x))/(n)`

Explanation:

Step 1: Define

Identify

`\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx`

Step 2: Integrate Pt. 1

Identify variables for u-substitution (trigonometric substitution).

1. Set u:
   `\displaystyle x = sin(u)`
2. [u] Differentiate [Trigonometric Differentiation]:
   `\displaystyle dx = cos(u) \ du`
3. Rewrite u:
   `\displaystyle u = arcsin(x)`

Step 3: Integrate Pt. 2

1. [Integral] Trigonometric Substitution:
   `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = \int\limits^a_b {cos(u)[1 - sin^2(u)]^\Big{(3)/(2)} \, du`
2. [Integrand] Rewrite:
   `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = \int\limits^a_b {cos(u)[cos^2(u)]^\Big{(3)/(2)} \, du`
3. [Integrand] Simplify:
   `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = \int\limits^a_b {cos^4(u)} \, du`
4. [Integral] Reduction Formula:
   `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = (4 - 1)/(4)\int \limits^a_b {cos^(4 - 2)(x)} \, dx + (cos^(4 - 1)(u)sin(u))/(4) \bigg| \limits^a_b`
5. [Integral] Simplify:
   `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = (cos^3(u)sin(u))/(4) \bigg| \limits^a_b + (3)/(4)\int\limits^a_b {cos^2(u)} \, du`
6. [Integral] Reduction Formula:
   `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = (cos^3(u)sin(u))/(4) \bigg|\limits^a_b + (3)/(4) \bigg[ (2 - 1)/(2)\int\limits^a_b {cos^(2 - 2)(u)} \, du + (cos^(2 - 1)(u)sin(u))/(2) \bigg| \limits^a_b \bigg]`
7. [Integral] Simplify:
   `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = (cos^3(u)sin(u))/(4) \bigg| \limits^a_b + (3)/(4) \bigg[ (1)/(2)\int\limits^a_b {} \, du + (cos(u)sin(u))/(2) \bigg| \limits^a_b \bigg]`
8. [Integral] Reverse Power Rule:
   `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = (cos^3(u)sin(u))/(4) \bigg| \limits^a_b + (3)/(4) \bigg[ (1)/(2)(u) \bigg| \limits^a_b + (cos(u)sin(u))/(2) \bigg| \limits^a_b \bigg]`
9. Simplify:
   `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = (cos^3(u)sin(u))/(4) \bigg| \limits^a_b + (3cos(u)sin(u))/(8) \bigg| \limits^a_b + (3)/(8)(u) \bigg| \limits^a_b`
10. Back-Substitute:
    `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = (cos^3(arcsin(x))sin(arcsin(x)))/(4) \bigg| \limits^a_b + (3cos(arcsin(x))sin(arcsin(x)))/(8) \bigg| \limits^a_b + (3)/(8)(arcsin(x)) \bigg| \limits^a_b`
11. Simplify:
    `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = (3arcsin(x))/(8) \bigg| \limits^a_b + \frac{x(1 - x^2)^\Big{(3)/(2)}}{4} \bigg| \limits^a_b + (3x√(1 - x^2))/(8) \bigg| \limits^a_b`
12. Rewrite:
    `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = \frac{3arcsin(x) + 2x(1 - x^2)^\Big{(3)/(2)} + 3x√(1 - x^2)}{8} \bigg| \limits^a_b`
13. Evaluate [Integration Rule - Fundamental Theorem of Calculus 1]:
    `\displaystyle \int\limits^a_b {(1 - x^2)^\Big{(3)/(2)}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{(3)/(2)} + 3a√(1 - a^2)}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{(3)/(2)} + 3b√(1 - b^2)}{8}`

Topic: AP Calculus AB/BC (Calculus I/I + II)

Unit: Integration

Book: College Calculus 10e