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Home/Calculus/Chapter 6/Computation Techniques

CALC-6.4

3-4 hours

Essential Skills

Computation Techniques

Master the art of computing definite integrals: substitution, integration by parts, symmetry tricks, and the famous Wallis and Stirling formulas.

Learning Objectives

Master substitution for definite integrals

Apply integration by parts with definite limits

Exploit symmetry properties (odd/even functions)

Handle periodic function integrals

Derive and apply the Wallis formula

Understand Stirling's approximation

Recognize common integral patterns

Solve challenging competition-level problems

1. Substitution Method

Theorem 6.19: Substitution for Definite Integrals

Let $f \in C[a, b]$ and $\phi \in C^1[\alpha, \beta]$ with $\phi(\alpha) = a$ , $\phi(\beta) = b$ , and $a \leq \phi(t) \leq b$ for all $t \in [\alpha, \beta]$ . Then:

\int_a^b f(x)\,dx = \int_\alpha^\beta f(\phi(t))\phi'(t)\,dt

Proof of Theorem 6.19:

Let $F$ be an antiderivative of $f$ . By the chain rule:

\frac{d}{dt}[F(\phi(t))] = F'(\phi(t))\phi'(t) = f(\phi(t))\phi'(t)

By Newton-Leibniz:

\int_\alpha^\beta f(\phi(t))\phi'(t)\,dt = F(\phi(t))\Big|_\alpha^\beta = F(\phi(\beta)) - F(\phi(\alpha)) = F(b) - F(a)

This equals $\int_a^b f(x)\,dx$ .

∎

Remark 6.7: Key Points for Substitution

1. Change the Limits!

When $x = \phi(t)$ , find new limits: if $x = a$ then $t = ?$ , if $x = b$ then $t = ?$ .

2. No Need to Back-Substitute

Unlike indefinite integrals, you evaluate at the new limits directly—no need to convert back to $x$ .

3. Limits Can Swap

If $\phi(\alpha) = b$ and $\phi(\beta) = a$ , the formula still works (with a sign change absorbed).

Example 6.17: Basic Substitution

Compute: $\int_0^2 x\sqrt{4 - x^2}\,dx$

Solution:

Let $u = 4 - x^2$ , so $du = -2x\,dx$ , i.e., $x\,dx = -\frac{1}{2}du$ .

When $x = 0$ : $u = 4$ . When $x = 2$ : $u = 0$ .

\int_0^2 x\sqrt{4 - x^2}\,dx = \int_4^0 \sqrt{u} \cdot \left(-\frac{1}{2}\right)du = \frac{1}{2}\int_0^4 \sqrt{u}\,du

= \frac{1}{2} \cdot \frac{2}{3}u^{3/2}\Big|_0^4 = \frac{1}{3} \cdot 8 = \frac{8}{3}

Example 6.18: Trigonometric Substitution

Compute: $\int_0^a \sqrt{a^2 - x^2}\,dx$

Solution:

Let $x = a\sin\theta$ , so $dx = a\cos\theta\,d\theta$ .

When $x = 0$ : $\theta = 0$ . When $x = a$ : $\theta = \pi/2$ .

\sqrt{a^2 - x^2} = \sqrt{a^2(1 - \sin^2\theta)} = a\cos\theta

\int_0^a \sqrt{a^2 - x^2}\,dx = \int_0^{\pi/2} a\cos\theta \cdot a\cos\theta\,d\theta = a^2 \int_0^{\pi/2} \cos^2\theta\,d\theta

Using $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ :

= a^2 \cdot \frac{1}{2} \left[\theta + \frac{\sin 2\theta}{2}\right]_0^{\pi/2} = a^2 \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi a^2}{4}

Geometric interpretation: This is the area of a quarter circle of radius $a$ !

Example 6.19: Clever Substitution

Compute: $\int_{-1}^2 \frac{x}{\sqrt{9 - 2x^2}}\,dx$

Solution:

Let $u = 9 - 2x^2$ , so $du = -4x\,dx$ .

When $x = -1$ : $u = 7$ . When $x = 2$ : $u = 1$ .

\int_{-1}^2 \frac{x\,dx}{\sqrt{9 - 2x^2}} = \int_7^1 \frac{-\frac{1}{4}du}{\sqrt{u}} = \frac{1}{4}\int_1^7 u^{-1/2}\,du

= \frac{1}{4} \cdot 2\sqrt{u}\Big|_1^7 = \frac{1}{2}(\sqrt{7} - 1)

2. Integration by Parts

Theorem 6.20: Integration by Parts for Definite Integrals

If $u, v \in C^1[a, b]$ , then:

\int_a^b u(x)v'(x)\,dx = u(x)v(x)\Big|_a^b - \int_a^b u'(x)v(x)\,dx

Or in differential notation:

\int_a^b u\,dv = uv\Big|_a^b - \int_a^b v\,du

Proof of Theorem 6.20:

By the product rule: $(uv)' = u'v + uv'$ .

Integrating both sides from $a$ to $b$ :

uv\Big|_a^b = \int_a^b u'v\,dx + \int_a^b uv'\,dx

Rearranging gives the formula.

∎

Example 6.20: Polynomial × Trigonometric

Compute: $\int_0^\pi x\sin x\,dx$

Solution:

Let $u = x$ , $dv = \sin x\,dx$ .

Then $du = dx$ , $v = -\cos x$ .

\int_0^\pi x\sin x\,dx = -x\cos x\Big|_0^\pi + \int_0^\pi \cos x\,dx

= [-\pi\cos\pi - 0] + \sin x\Big|_0^\pi = \pi + 0 = \pi

Example 6.21: Logarithmic Function

Compute: $\int_1^e x\ln x\,dx$

Solution:

Let $u = \ln x$ , $dv = x\,dx$ .

Then $du = \frac{1}{x}dx$ , $v = \frac{x^2}{2}$ .

\int_1^e x\ln x\,dx = \frac{x^2}{2}\ln x\Big|_1^e - \int_1^e \frac{x^2}{2} \cdot \frac{1}{x}\,dx

= \frac{e^2}{2} \cdot 1 - 0 - \frac{1}{2}\int_1^e x\,dx = \frac{e^2}{2} - \frac{1}{2} \cdot \frac{x^2}{2}\Big|_1^e

= \frac{e^2}{2} - \frac{1}{4}(e^2 - 1) = \frac{e^2}{4} + \frac{1}{4} = \frac{e^2 + 1}{4}

Example 6.22: Higher Powers

Compute: $\int_1^e x^2(\ln x)^2\,dx$

Solution:

Let $u = (\ln x)^2$ , $dv = x^2\,dx$ .

Then $du = \frac{2\ln x}{x}dx$ , $v = \frac{x^3}{3}$ .

\int_1^e x^2(\ln x)^2\,dx = \frac{x^3}{3}(\ln x)^2\Big|_1^e - \frac{2}{3}\int_1^e x^2\ln x\,dx

For the remaining integral, use parts again with $u = \ln x$ , $dv = x^2\,dx$ :

\int_1^e x^2\ln x\,dx = \frac{x^3}{3}\ln x\Big|_1^e - \frac{1}{3}\int_1^e x^2\,dx = \frac{e^3}{3} - \frac{1}{9}(e^3 - 1)

= \frac{2e^3}{9} + \frac{1}{9}

Substituting back:

= \frac{e^3}{3} - \frac{2}{3}\left(\frac{2e^3}{9} + \frac{1}{9}\right) = \frac{e^3}{3} - \frac{4e^3}{27} - \frac{2}{27} = \frac{5e^3 - 2}{27}

Theorem 6.21: Generalized Integration by Parts

If $f, g \in R[a, b]$ and $F(x) = \int_a^x f(t)\,dt$ , $G(x) = \int_a^x g(t)\,dt$ , then:

\int_a^b F(x)g(x)\,dx = F(x)G(x)\Big|_a^b - \int_a^b G(x)f(x)\,dx

This version works even when $f, g$ are only integrable (not necessarily continuous).

3. Symmetry Properties

Definition 6.9: Odd and Even Functions

Even Function

f(-x) = f(x)

Symmetric about the y-axis

Odd Function

f(-x) = -f(x)

Symmetric about the origin

Theorem 6.22: Symmetry Integration Rules

For $f \in C[-a, a]$ :

If $f$ is even:

\int_{-a}^a f(x)\,dx = 2\int_0^a f(x)\,dx

If $f$ is odd:

\int_{-a}^a f(x)\,dx = 0

Proof of Theorem 6.22:

Split the integral and substitute $u = -x$ in the left part:

\int_{-a}^a f(x)\,dx = \int_{-a}^0 f(x)\,dx + \int_0^a f(x)\,dx

For the first integral, let $u = -x$ :

\int_{-a}^0 f(x)\,dx = \int_a^0 f(-u)(-du) = \int_0^a f(-u)\,du

If $f$ is even: $f(-u) = f(u)$ , so total = $2\int_0^a f(x)\,dx$ .

If $f$ is odd: $f(-u) = -f(u)$ , so total = 0.

∎

Example 6.23: Using Symmetry

Compute: $\int_{-1}^1 (x^3 + x^2 + x + 1)\,dx$

Solution:

Split into odd and even parts:

Odd: $x^3 + x$ → integral = 0
Even: $x^2 + 1$ → use doubling

\int_{-1}^1 (x^3 + x^2 + x + 1)\,dx = 0 + 2\int_0^1 (x^2 + 1)\,dx = 2\left[\frac{x^3}{3} + x\right]_0^1 = 2 \cdot \frac{4}{3} = \frac{8}{3}

Example 6.24: Hidden Symmetry

Compute: $\int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx$

Solution:

Let $I = \int_0^{\pi/2} \frac{\sin x}{\sin x + \cos x}\,dx$ .

Substitute $u = \frac{\pi}{2} - x$ :

I = \int_{\pi/2}^0 \frac{\sin(\frac{\pi}{2} - u)}{\sin(\frac{\pi}{2} - u) + \cos(\frac{\pi}{2} - u)}(-du) = \int_0^{\pi/2} \frac{\cos u}{\cos u + \sin u}\,du

Adding the two expressions for $I$ :

2I = \int_0^{\pi/2} \frac{\sin x + \cos x}{\sin x + \cos x}\,dx = \int_0^{\pi/2} 1\,dx = \frac{\pi}{2}

Therefore: $I = \frac{\pi}{4}$

Theorem 6.23: King's Property

For $f \in C[a, b]$ :

\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx

This reflects the interval about its midpoint $\frac{a+b}{2}$ .

Corollary 6.6: Classic Application

For continuous $f$ on $[0, a]$ :

\int_0^a \frac{f(x)}{f(x) + f(a-x)}\,dx = \frac{a}{2}

Proof of Corollary 6.6:

Let $I = \int_0^a \frac{f(x)}{f(x) + f(a-x)}\,dx$ .

Substitute $u = a - x$ :

I = \int_a^0 \frac{f(a-u)}{f(a-u) + f(u)}(-du) = \int_0^a \frac{f(a-u)}{f(a-u) + f(u)}\,du

Adding:

2I = \int_0^a \frac{f(x) + f(a-x)}{f(x) + f(a-x)}\,dx = \int_0^a 1\,dx = a

∎

4. Periodic Function Integrals

Definition 6.10: Periodic Function

A function $f$ is periodic with period $T > 0$ if:

f(x + T) = f(x) \quad \text{for all } x

Theorem 6.24: Period Integral Property

If $f$ is continuous and periodic with period $T$ , then for any $a \in \mathbb{R}$ :

\int_a^{a+T} f(x)\,dx = \int_0^T f(x)\,dx

Proof of Theorem 6.24:

\int_a^{a+T} f(x)\,dx = \int_a^0 f(x)\,dx + \int_0^T f(x)\,dx + \int_T^{a+T} f(x)\,dx

In the last integral, let $u = x - T$ :

\int_T^{a+T} f(x)\,dx = \int_0^a f(u+T)\,du = \int_0^a f(u)\,du

So $\int_a^0 + \int_T^{a+T} = -\int_0^a + \int_0^a = 0$ .

∎

Example 6.25: Periodic Integral

Compute: $\int_\pi^{3\pi} \sin^2 x\,dx$

Solution:

$\sin^2 x$ has period $\pi$ . The interval $[\pi, 3\pi]$ spans 2 periods.

\int_\pi^{3\pi} \sin^2 x\,dx = 2\int_0^\pi \sin^2 x\,dx = 2 \cdot \frac{\pi}{2} = \pi

5. The Wallis Formula

Definition 6.11: Wallis Integral

Define the Wallis integral:

I_n = \int_0^{\pi/2} \sin^n x\,dx = \int_0^{\pi/2} \cos^n x\,dx

Theorem 6.25: Wallis Reduction Formula

For $n \geq 2$ :

I_n = \frac{n-1}{n} I_{n-2}

Proof of Theorem 6.25:

Use integration by parts with $u = \sin^{n-1}x$ , $dv = \sin x\,dx$ :

I_n = \int_0^{\pi/2} \sin^{n-1}x \cdot \sin x\,dx = -\sin^{n-1}x\cos x\Big|_0^{\pi/2} + (n-1)\int_0^{\pi/2} \sin^{n-2}x \cos^2 x\,dx

The boundary term is 0. Using $\cos^2 x = 1 - \sin^2 x$ :

I_n = (n-1)\int_0^{\pi/2} \sin^{n-2}x(1 - \sin^2 x)\,dx = (n-1)I_{n-2} - (n-1)I_n

Solving for $I_n$ : $nI_n = (n-1)I_{n-2}$ .

∎

Corollary 6.7: Explicit Formulas

With $I_0 = \frac{\pi}{2}$ and $I_1 = 1$ :

Even index:

I_{2n} = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}

Odd index:

I_{2n+1} = \frac{(2n)!!}{(2n+1)!!}

Here $n!! = n(n-2)(n-4)\cdots$ is the double factorial.

Example 6.26: Computing Wallis Integrals

Compute: $I_4$ and $I_5$

Solution:

For $I_4$ :

I_4 = \frac{3}{4}I_2 = \frac{3}{4} \cdot \frac{1}{2}I_0 = \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{3\pi}{16}

For $I_5$ :

I_5 = \frac{4}{5}I_3 = \frac{4}{5} \cdot \frac{2}{3}I_1 = \frac{4}{5} \cdot \frac{2}{3} \cdot 1 = \frac{8}{15}

Theorem 6.26: Wallis Product for π

The Wallis formula gives a remarkable product representation:

\frac{\pi}{2} = \lim_{n \to \infty} \frac{(2n)!!^2}{(2n-1)!!^2(2n+1)} = \prod_{k=1}^\infty \frac{4k^2}{4k^2 - 1}

Proof of Theorem 6.26:

For $x \in (0, \pi/2)$ : $\sin^{2n+1}x < \sin^{2n}x < \sin^{2n-1}x$ .

Integrating: $I_{2n+1} < I_{2n} < I_{2n-1}$ .

1 < \frac{I_{2n}}{I_{2n+1}} < \frac{I_{2n-1}}{I_{2n+1}} = \frac{2n+1}{2n} \to 1

By squeeze theorem, $\frac{I_{2n}}{I_{2n+1}} \to 1$ . Expanding:

\frac{I_{2n}}{I_{2n+1}} = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2} \cdot \frac{(2n+1)!!}{(2n)!!} \to 1

Rearranging gives the Wallis product.

∎

6. Stirling's Approximation

Theorem 6.27: Stirling's Formula

As $n \to \infty$ :

n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n

More precisely: $n! = \sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{\theta_n/(12n)}$ where $0 < \theta_n < 1$ .

Proof of Theorem 6.27 (Sketch):

Define $a_n = \frac{n! e^n}{n^{n+1/2}}$ . We show $a_n \to \sqrt{2\pi}$ .

Step 1: Show $\{a_n\}$ is decreasing and bounded below.

\frac{a_n}{a_{n+1}} = \frac{1}{e}\left(1 + \frac{1}{n}\right)^{n+1/2}

Using Hadamard's inequality: $1 \leq \frac{a_n}{a_{n+1}} \leq e^{1/(4n(n+1))}$ .

Step 2: Let $\alpha = \lim a_n$ . Show $\alpha > 0$ .

Step 3: Use Wallis formula to find $\alpha = \sqrt{2\pi}$ :

\sqrt{\pi} = \lim_{n\to\infty} \frac{(2n)!!}{(2n-1)!!\sqrt{n}} = \lim_{n\to\infty} \frac{(n!)^2 2^{2n}}{(2n)!\sqrt{n}} = \frac{\alpha^2}{\alpha\sqrt{2}}

Solving: $\alpha = \sqrt{2\pi}$ .

∎

Example 6.27: Using Stirling's Formula

Estimate: $10!$

Solution:

10! \approx \sqrt{2\pi \cdot 10} \cdot \left(\frac{10}{e}\right)^{10} \approx \sqrt{62.83} \cdot 3.679^{10}

\approx 7.926 \times 453,999 \approx 3,598,696

Exact value: $10! = 3,628,800$ . Error ≈ 0.8%.

Remark 6.8: Applications of Stirling

Combinatorics

Estimate binomial coefficients, counting problems

Probability

Central limit theorem, random walks

Statistical Mechanics

Entropy calculations, partition functions

Algorithm Analysis

Complexity of sorting, permutation algorithms

7. Advanced Integration Techniques

Example 6.28: Taylor Formula via Integration

Derive: Taylor's formula with integral remainder using integration by parts.

Derivation:

Start with $f(x) - f(x_0) = \int_{x_0}^x f'(t)\,dt$ .

Use parts with $u = f'(t)$ , $dv = dt = d(t-x)$ :

= f'(t)(t-x)\Big|_{x_0}^x - \int_{x_0}^x f''(t)(t-x)\,dt

= f'(x_0)(x-x_0) + \int_{x_0}^x f''(t)(x-t)\,dt

Continuing this process $n$ times:

f(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x-x_0)^k + \frac{1}{n!}\int_{x_0}^x f^{(n+1)}(t)(x-t)^n\,dt

Example 6.29: Dirichlet Integral (Preview)

Famous result:

\int_0^\infty \frac{\sin x}{x}\,dx = \frac{\pi}{2}

This is an improper integral (covered in Chapter 7). The proof uses clever techniques including Laplace transforms or Feynman's differentiation under the integral.

Example 6.30: Iterated Integral

Prove: $\int_a^x \left(\int_a^t f(u)\,du\right)dt = \int_a^x (x-t)f(t)\,dt$

Proof:

Let $F(t) = \int_a^t f(u)\,du$ . Use parts with $u = F(t)$ , $dv = dt$ :

\int_a^x F(t)\,dt = \int_a^x F(t)\,d(t-a) = F(t)(t-a)\Big|_a^x - \int_a^x (t-a)f(t)\,dt

Actually, use $dv = d(t-x)$ instead:

= F(t)(t-x)\Big|_a^x - \int_a^x (t-x)f(t)\,dt = 0 + \int_a^x (x-t)f(t)\,dt

Example 6.31: Logarithmic Integral

Compute: $\int_0^{\pi/2} \ln(\sin x)\,dx$

Solution:

Let $I = \int_0^{\pi/2} \ln(\sin x)\,dx$ . By symmetry with $u = \pi/2 - x$ :

I = \int_0^{\pi/2} \ln(\cos x)\,dx

Adding:

2I = \int_0^{\pi/2} \ln(\sin x \cos x)\,dx = \int_0^{\pi/2} \ln\left(\frac{\sin 2x}{2}\right)dx

= \int_0^{\pi/2} \ln(\sin 2x)\,dx - \frac{\pi}{2}\ln 2

Let $t = 2x$ in the first integral:

\int_0^{\pi/2} \ln(\sin 2x)\,dx = \frac{1}{2}\int_0^\pi \ln(\sin t)\,dt = \frac{1}{2} \cdot 2I = I

So: $2I = I - \frac{\pi}{2}\ln 2$ , giving $I = -\frac{\pi}{2}\ln 2$ .

Example 6.32: Reduction Formula Application

Compute: $\int_0^{\pi} x^2\sin x\,dx$

Solution:

Use parts twice. Let $u = x^2$ , $dv = \sin x\,dx$ :

\int_0^{\pi} x^2\sin x\,dx = -x^2\cos x\Big|_0^\pi + 2\int_0^\pi x\cos x\,dx

= \pi^2 + 2\left[x\sin x\Big|_0^\pi - \int_0^\pi \sin x\,dx\right]

= \pi^2 + 2\left[0 + \cos x\Big|_0^\pi\right] = \pi^2 + 2(-1-1) = \pi^2 - 4

Example 6.33: Gamma Function Preview

The Gamma Function:

\Gamma(n) = \int_0^\infty x^{n-1}e^{-x}\,dx = (n-1)!

For positive integers, this equals the factorial. The Gamma function extends factorials to all complex numbers (except non-positive integers). The key property is $\Gamma(n+1) = n\Gamma(n)$ .

Example 6.34: Beta Function Preview

The Beta Function:

B(m, n) = \int_0^1 x^{m-1}(1-x)^{n-1}\,dx = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}

Special case: $B(1/2, 1/2) = \pi$ . The Wallis integral $I_n$ can be expressed using Beta functions.

Example 6.35: Fresnel Integrals

Fresnel Integrals:

C(x) = \int_0^x \cos\left(\frac{\pi t^2}{2}\right)dt

S(x) = \int_0^x \sin\left(\frac{\pi t^2}{2}\right)dt

These integrals arise in optics (diffraction patterns) and have the remarkable limits: $C(\infty) = S(\infty) = \frac{1}{2}$

8. Practice Problems

Problem Set A: Basic Techniques

1. $\int_0^4 \frac{x}{\sqrt{1+2x}}\,dx$

2. $\int_0^{\pi/4} \tan^4 x\,dx$

3. $\int_0^1 x\arctan x\,dx$

4. $\int_1^e \frac{\ln x}{x^2}\,dx$

5. $\int_0^{\pi} x\cos^2 x\,dx$

6. $\int_0^1 x^2 e^{-x}\,dx$

Problem Set B: Symmetry Tricks

7. $\int_{-\pi}^{\pi} \frac{x\sin x}{1+\cos^2 x}\,dx$

8. $\int_0^{\pi/2} \frac{\cos^3 x}{\sin^3 x + \cos^3 x}\,dx$

9. $\int_0^1 \frac{\ln(1+x)}{1+x^2}\,dx$

10. $\int_0^\pi \frac{x\sin x}{1+\cos^2 x}\,dx$

Problem Set C: Wallis-Type Integrals

11. $\int_0^{\pi/2} \sin^6 x\,dx$

12. $\int_0^{\pi/2} \sin^3 x \cos^4 x\,dx$

13. $\int_0^{\pi/2} \sin^2 x \cos^4 x\,dx$

14. $\int_0^\pi \sin^5 x\,dx$

Problem Set D: Challenge Problems

15. $\lim_{n\to\infty} \int_0^1 \frac{nx^{n-1}}{1+x}\,dx$

16. $\int_0^{\pi/4} \ln(1+\tan x)\,dx$

17. Prove $I_{2n}I_{2n+1} = \frac{\pi}{2(2n+1)}$

18. $\int_0^1 \frac{\ln x}{1-x^2}\,dx$

Remark 6.9: Answers to Selected Problems

\frac{26}{9}

\frac{\pi}{4} - \frac{1}{2}

\frac{\pi}{4}

10:

\frac{\pi^2}{4}

11:

\frac{5\pi}{32}

12:

\frac{2}{35}

14:

\frac{16}{15}

16:

\frac{\pi\ln 2}{8}

9. Formula Summary

Technique	Formula	When to Use
Substitution	$\int_a^b f(\phi(t))\phi'(t)\,dt$	Composite functions, radicals
Parts	$\int u\,dv = uv\|_a^b - \int v\,du$	Products of different types
Even symmetry	$\int_{-a}^a f = 2\int_0^a f$	$f(-x) = f(x)$
Odd symmetry	$\int_{-a}^a f = 0$	$f(-x) = -f(x)$
King's property	$\int_a^b f(x) = \int_a^b f(a+b-x)$	Midpoint reflection
Periodic	$\int_a^{a+T} f = \int_0^T f$	Period $T$
Wallis	$I_n = \frac{n-1}{n}I_{n-2}$	$\sin^n x, \cos^n x$
Stirling	$n! \sim \sqrt{2\pi n}(n/e)^n$	Large factorials

10. Common Mistakes to Avoid

❌ Forgetting to change limits

When substituting, ALWAYS update the limits of integration!

❌ Wrong boundary in parts

Don't forget to evaluate $uv|_a^b$ at BOTH limits!

❌ Misidentifying odd/even

$e^{-x^2}$ is even, but $xe^{-x^2}$ is odd!

❌ Wrong Wallis recursion

It's $I_n = \frac{n-1}{n}I_{n-2}$ , not $\frac{n}{n-1}$ !

❌ Ignoring domain issues

Check that substitutions are valid on the integration interval!

❌ Sign errors in symmetry

When $u = -x$ , don't forget $du = -dx$ !

11. Historical Notes

John Wallis (1616-1703)

English mathematician who discovered the product formula for π in 1656. His work “Arithmetica Infinitorum” influenced Newton's development of calculus.

\frac{\pi}{2} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots

James Stirling (1692-1770)

Scottish mathematician who refined the factorial approximation. Though de Moivre found it first, Stirling determined the constant $\sqrt{2\pi}$ .

n! \sim \sqrt{2\pi n} \left(\frac{n}{e}\right)^n

Augustin-Louis Cauchy (1789-1857)

French mathematician who rigorized calculus. The Cauchy-Schwarz inequality he proved is fundamental in analysis, probability, and physics.

\left(\int fg\right)^2 \leq \int f^2 \cdot \int g^2

Leonhard Euler (1707-1783)

Swiss mathematician who discovered countless integral formulas and introduced the Gamma function, extending factorials to non-integers.

\Gamma(n) = (n-1)! = \int_0^\infty x^{n-1}e^{-x}\,dx

12. Connections to Other Topics

Probability Theory

The Gaussian integral $\int e^{-x^2}dx = \sqrt{\pi}$ is the foundation of the normal distribution in statistics.

Fourier Analysis

Wallis integrals appear in computing Fourier coefficients of periodic functions, essential for signal processing.

Physics

Stirling's formula is crucial in statistical mechanics for calculating entropy and partition functions.

Complex Analysis

Many definite integrals are evaluated using contour integration in the complex plane, a powerful technique from complex analysis.

Number Theory

The Riemann zeta function $\zeta(s) = \sum n^{-s}$ connects to integrals and the distribution of prime numbers.

Numerical Methods

Understanding exact integration helps design and analyze numerical integration methods like Simpson's rule and Gaussian quadrature.

Computation Techniques Quiz

Questions

Correct

Accuracy

For

\int_a^b f(x)\,dx

with substitution

x = \phi(t)

, the new limits are:

Easy

Not attempted

For odd function

f

[-a, a]

\int_{-a}^a f(x)\,dx

equals:

Easy

Not attempted

For even function

f

[-a, a]

\int_{-a}^a f(x)\,dx

equals:

Easy

Not attempted

The Wallis formula gives

I_n = \int_0^{\pi/2} \sin^n x\,dx

as:

Medium

Not attempted

\int_0^{\pi/2} \sin^4 x\,dx

equals:

Medium

Not attempted

Stirling's approximation states

n! \sim

Medium

Not attempted

\int_0^a \frac{f(x)}{f(x) + f(a-x)}\,dx

equals:

Hard

Not attempted

For periodic

f

with period

T

\int_a^{a+T} f(x)\,dx

Medium

Not attempted

\int_0^\pi x\sin x\,dx

using parts gives:

Medium

Not attempted

The integral

\int_0^{\pi/2} \ln(\sin x)\,dx

equals:

Hard

Not attempted

Frequently Asked Questions

When should I use substitution vs. parts?

Use substitution when you see a composite function f(g(x))g'(x). Use parts when you have a product of different function types (polynomial × exponential, polynomial × trig, etc.).

How do I change limits in substitution?

If x = φ(t), when x = a find t = φ⁻¹(a), and when x = b find t = φ⁻¹(b). These become your new limits.

Why does the odd function integral vanish?

Geometrically, the negative area on [-a, 0] exactly cancels the positive area on [0, a] due to point symmetry about the origin.

What makes Wallis formula so useful?

It reduces powers of sin/cos to simpler integrals via recursion, and leads to remarkable results like the Wallis product for π.

How accurate is Stirling's approximation?

The relative error is about 1/(12n), so for n = 10 it's ~1% accurate, for n = 100 it's ~0.1% accurate.

What's the trick for ∫f(x)/(f(x)+f(a-x))?

Substitute u = a - x to get an equivalent integral, then add both versions. The sum simplifies to the integral of 1, giving a/2.

Key Takeaways

Substitution

Change limits when substituting!

x = \phi(t) \Rightarrow \text{new limits}

Parts Formula

\int u\,dv = uv|_a^b - \int v\,du

Odd Function

\int_{-a}^a f_{\text{odd}} = 0

Even Function

\int_{-a}^a f_{\text{even}} = 2\int_0^a f

King's Property

\int_a^b f(x) = \int_a^b f(a+b-x)

Wallis Reduction

I_n = \frac{n-1}{n}I_{n-2}

Periodic

\int_a^{a+T} f = \int_0^T f

Stirling

n! \sim \sqrt{2\pi n}(n/e)^n

Classic Trick

\int_0^a \frac{f}{f+f(a-x)} = \frac{a}{2}

Quick Reference Tips

Trigonometric Powers

For $\int \sin^m x \cos^n x\,dx$ :

If $m$ odd: save one $\sin x$ , use $\sin^2 = 1 - \cos^2$
If $n$ odd: save one $\cos x$ , use $\cos^2 = 1 - \sin^2$
If both even: use half-angle formulas

Substitution Hints

Common substitutions:

$\sqrt{a^2-x^2}$ : $x = a\sin\theta$
$\sqrt{a^2+x^2}$ : $x = a\tan\theta$
$\sqrt{x^2-a^2}$ : $x = a\sec\theta$

Parts Strategy (LIATE)

Choose $u$ in this order:

Logarithmic: $\ln x, \log x$
Inverse trig: $\arcsin, \arctan$
Algebraic: $x^n, x^2+1$
Trigonometric: $\sin, \cos$
Exponential: $e^x, 2^x$