Question

rmula1" title="\int\limits^a_b {(1-x^{2} )^{3/2} } \, dx" alt="\int\limits^a_b {(1-x^{2} )^{3/2} } \, dx" align="absmiddle" class="latex-formula">

Answer 1

Answer:$\displaystyle \int\limits^a_b {(1 - x^2)^\Big{\frac{3}{2}}} \, dx = \frac{3arcsin(a) + 2a(1 - a^2)^\Big{\frac{3}{2}} + 3a\sqrt{1 - a^2}}{8} - \frac{3arcsin(b) + 2b(1 - b^2)^\Big{\frac{3}{2}} + 3b\sqrt{1 - b^2}}{8}$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]:                                                               $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

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:                                                                                               $\displaystyle \int {cos^n(x)} \, dx = \frac{n - 1}{n}\int {cos^{n - 2}(x)} \, dx + \frac{cos^{n - 1}(x)sin(x)}{n}$

Step-by-step explanation:

Step 1: Define

Identify

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

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

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

Unit: Integration

Book: College Calculus 10e

Answer 2

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,

$\int\cos^m(u)du=\frac{1}{m}\sin(u)\cos^{m-1}(u)+\frac{m-1}{m}\int\cos^{m-2}(u)du$

to get,

$\int\cos^4(u)du=\frac{1}{4}\sin(u)\cos^3(u)+\frac{3}{4}\int\cos^2(u)du$

Notice that,

$\cos^2(u)=\frac{1}{2}(\cos(2u)+1)$

Then integrate the obtained sum,

$\frac{1}{4}\sin(u)\cos^3(u)+\frac{3}{8}\int\cos(2u)du+\frac{3}{8}\int1du$

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

$\frac{3\sin(s)}{16}+\frac{1}{4}\sin(u)\cos^3(u)+\frac{3}{8}\int1du$

$\frac{3\sin(s)}{16}+\frac{3u}{4}+\frac{1}{4}\sin(u)\cos^3(u)+C$

Substitute 2u back for s,

$\frac{3u}{8}+\frac{1}{4}\sin(u)\cos^3(u)+\frac{3}{8}\sin(u)\cos(u)+C$

Substitute $\sin^{-1}$ for u and simplify with $\cos(\arcsin(x))=\sqrt{1-x^2}$ to get the result,

$\boxed{\frac{1}{8}(x\sqrt{1-x^2}(5-2x^2)+3\arcsin(x))+C}$

Let $F(x)=\frac{1}{8}(x\sqrt{1-x^2}(5-2x^2)+3\arcsin(x))+C$

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

$F(x)\Big|_b^a=F(a)-F(b)=\boxed{\frac{1}{8}(a\sqrt{1-a^2}(5-2a^2)+3\arcsin(a))-\frac{1}{8}(b\sqrt{1-b^2}(5-2b^2)+3\arcsin(b))}$

Hope this helps :)