Techniques of Integration II: Integration by Parts and Trigonometric Integrals

TL;DR

Integration by Parts helps integrate products of functions by transforming them into easier integrals. Trigonometric integrals use identities to simplify expressions involving powers of trigonometric functions. Mastering these techniques expands your ability to solve a wider range of complex integrals.

1. The Mental Model

Think of integration by parts as a strategic trade: you're swapping a difficult integral for a potentially simpler one by cleverly differentiating one part and integrating another. For trigonometric integrals, it's like having a toolkit of identities to reshape a complicated expression into something solvable.

2. The Core Material

Integration by Parts (IBP)

Integration by Parts is derived from the product rule for differentiation. If you have two functions, $u(x)$ and $v(x)$, then:
$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$

Integrating both sides with respect to $x$:
$\int \frac{d}{dx}(uv) dx = \int u\frac{dv}{dx} dx + \int v\frac{du}{dx} dx$
$uv = \int u \, dv + \int v \, du$

Rearranging this gives the integration by parts formula:
$\int u \, dv = uv - \int v \, du$

The key is choosing which part of your integrand becomes $u$ and which becomes $dv$. A helpful mnemonic for choosing $u$ is LIATE:
- Logarithmic functions (e.g., $\ln x$)
- Inverse trigonometric functions (e.g., $\arctan x$)
- Algebraic functions (e.g., $x^n$)
- Trigonometric functions (e.g., $\sin x$, $\cos x$)
- Exponential functions (e.g., $e^x$)

You generally want to choose $u$ as the function that comes first in the LIATE order, because it tends to simplify when differentiated. The $dv$ part should be something you can easily integrate.

graph TD
    Start["Given integral ∫ f(x) dx"] --> ChooseUDV["Choose u and dv from f(x) dx"]
    ChooseUDV --> Criteria["Criteria for Choosing u: LIATE mnemonic"]
    Criteria --> GoodU["u should simplify when differentiated"]
    Criteria --> GoodDV["dv should be easily integrable"]
    GoodU --> CalculateDU["Calculate du = u' dx"]
    GoodDV --> CalculateV["Integrate dv to find v"]
    CalculateDU & CalculateV --> ApplyFormula["Apply ∫ u dv = uv - ∫ v du"]
    ApplyFormula --> EvaluateNewIntegral["Evaluate the new integral ∫ v du"]
    EvaluateNewIntegral --> Done["Result of the integral"]

Trigonometric Integrals

These involve powers of sine, cosine, tangent, secant, etc. The strategy is to use trigonometric identities to transform the integrand into a form that's easier to integrate.

Integrals of Powers of Sine and Cosine ($\int \sin^m x \cos^n x \, dx$)

1. If n is odd: Save one $\cos x$ and convert the remaining $\cos^{n-1} x$ to powers of $\sin x$ using $\cos^2 x = 1 - \sin^2 x$. Let $u = \sin x$.
   Example: $\int \sin^2 x \cos^3 x \, dx = \int \sin^2 x \cos^2 x \cos x \, dx = \int \sin^2 x (1 - \sin^2 x) \cos x \, dx$. Let $u = \sin x$, $du = \cos x \, dx$.
   $\int u^2 (1 - u^2) \, du = \int (u^2 - u^4) \, du$.

2. If m is odd: Save one $\sin x$ and convert the remaining $\sin^{m-1} x$ to powers of $\cos x$ using $\sin^2 x = 1 - \cos^2 x$. Let $u = \cos x$.
   Example: $\int \sin^3 x \cos^2 x \, dx = \int \sin^2 x \cos^2 x \sin x \, dx = \int (1 - \cos^2 x) \cos^2 x \sin x \, dx$. Let $u = \cos x$, $du = -\sin x \, dx$.
   $\int (1 - u^2) u^2 (-du) = \int (u^4 - u^2) \, du$.

3. If both m and n are even: Use half-angle identities:
   $\sin^2 x = \frac{1 - \cos(2x)}{2}$
   $\cos^2 x = \frac{1 + \cos(2x)}{2}$
   $\sin x \cos x = \frac{1}{2} \sin(2x)$
   Repeatedly apply these until the powers are gone or you have an odd power of sine or cosine.

Integrals of Powers of Tangent and Secant ($\int \tan^m x \sec^n x \, dx$)

1. If n is even (and $n \ge 2$): Save one $\sec^2 x$ and convert the remaining $\sec^{n-2} x$ to powers of $\tan x$ using $\sec^2 x = 1 + \tan^2 x$. Let $u = \tan x$.
   Example: $\int \tan^2 x \sec^4 x \, dx = \int \tan^2 x \sec^2 x \sec^2 x \, dx = \int \tan^2 x (1 + \tan^2 x) \sec^2 x \, dx$. Let $u = \tan x$, $du = \sec^2 x \, dx$.
   $\int u^2 (1 + u^2) \, du = \int (u^2 + u^4) \, du$.

2. If m is odd (and $m \ge 1$): Save one $\sec x \tan x$ and convert the remaining $\tan^{m-1} x$ to powers of $\sec x$ using $\tan^2 x = \sec^2 x - 1$. Let $u = \sec x$.
   Example: $\int \tan^3 x \sec x \, dx = \int \tan^2 x \sec x \tan x \, dx = \int (\sec^2 x - 1) \sec x \tan x \, dx$. Let $u = \sec x$, $du = \sec x \tan x \, dx$.
   $\int (u^2 - 1) \, du$.

3. Other cases: Sometimes you'll need to use identities or even IBP. For instance, $\int \tan x \, dx = \ln|\sec x| + C$, and $\int \sec x \, dx = \ln|\sec x + \tan x| + C$. For even powers of $\tan x$ and odd powers of $\sec x$, it can be more complex, sometimes requiring reduction formulas.

3. Worked Example

Let's evaluate $\int x e^{2x} \, dx$ using Integration by Parts and then $\int \sin^4 x \, dx$ using trigonometric identities.

Part 1: $\int x e^{2x} \, dx$ (Integration by Parts)

1. Choose $u$ and $dv$: Using LIATE, Algebraic ($x$) comes before Exponential ($e^{2x}$).
   Let $u = x$
   Let $dv = e^{2x} \, dx$

2. Calculate $du$ and $v$:
   $du = dx$
   To find $v$, integrate $dv$: $\int e^{2x} \, dx = \frac{1}{2}e^{2x}$. So, $v = \frac{1}{2}e^{2x}$.

3. Apply the IBP formula: $\int u \, dv = uv - \int v \, du$
   $\int x e^{2x} \, dx = x \left(\frac{1}{2}e^{2x}\right) - \int \left(\frac{1}{2}e^{2x}\right) dx$
   $\int x e^{2x} \, dx = \frac{1}{2}x e^{2x} - \frac{1}{2} \int e^{2x} \, dx$

4. Evaluate the new integral:
   $\int e^{2x} \, dx = \frac{1}{2}e^{2x}$

5. Substitute back and add the constant of integration:
   $\int x e^{2x} \, dx = \frac{1}{2}x e^{2x} - \frac{1}{2} \left(\frac{1}{2}e^{2x}\right) + C$
   $\int x e^{2x} \, dx = \frac{1}{2}x e^{2x} - \frac{1}{4}e^{2x} + C$

Part 2: $\int \sin^4 x \, dx$ (Trigonometric Integral)

Here, we have an even power of sine. We need to use half-angle identities.

1. Rewrite using $\sin^2 x$:
   $\int \sin^4 x \, dx = \int (\sin^2 x)^2 \, dx$

2. Apply the half-angle identity $\sin^2 x = \frac{1 - \cos(2x)}{2}$:
   $\int \left(\frac{1 - \cos(2x)}{2}\right)^2 \, dx = \int \frac{1 - 2\cos(2x) + \cos^2(2x)}{4} \, dx$
   $= \frac{1}{4} \int (1 - 2\cos(2x) + \cos^2(2x)) \, dx$

3. Apply another half-angle identity for $\cos^2(2x)$: Use $\cos^2 \theta = \frac{1 + \cos(2\theta)}{2}$. Here, $\theta = 2x$, so $2\theta = 4x$.
   $\cos^2(2x) = \frac{1 + \cos(4x)}{2}$

4. Substitute back into the integral:
   $= \frac{1}{4} \int \left(1 - 2\cos(2x) + \frac{1 + \cos(4x)}{2}\right) \, dx$
   $= \frac{1}{4} \int \left(1 - 2\cos(2x) + \frac{1}{2} + \frac{1}{2}\cos(4x)\right) \, dx$
   $= \frac{1}{4} \int \left(\frac{3}{2} - 2\cos(2x) + \frac{1}{2}\cos(4x)\right) \, dx$

5. Integrate term by term:
   $= \frac{1}{4} \left[\frac{3}{2}x - 2\left(\frac{\sin(2x)}{2}\right) + \frac{1}{2}\left(\frac{\sin(4x)}{4}\right)\right] + C$
   $= \frac{1}{4} \left[\frac{3}{2}x - \sin(2x) + \frac{1}{8}\sin(4x)\right] + C$
   $= \frac{3}{8}x - \frac{1}{4}\sin(2x) + \frac{1}{32}\sin(4x) + C$

4. Key Takeaways

• Integration by Parts is essential for integrating products of functions that don't simplify with basic substitution.
• The LIATE mnemonic is a helpful guide for choosing $u$ in IBP, aiming for $u$ to simplify upon differentiation.
• Trigonometric identities are your main tools for simplifying trigonometric integrals, often reducing them to solvable polynomial or simpler trigonometric forms.
• For $\sin^m x \cos^n x$, look for odd powers first to use simple $u$-substitution.
• For $\tan^m x \sec^n x$, look for even powers of secant or odd powers of tangent.
• Half-angle identities are crucial when dealing with even powers of sine or cosine.

Common Mistakes to Avoid:
- Incorrectly choosing $u$ and $dv$ in IBP: A poor choice can lead to a more complex integral than you started with. Always check if $\int v \, du$ is simpler.
- Forgetting to integrate $dv$ to get $v$: This is a fundamental step in IBP.
- Algebra errors with trigonometric identities: Double-check your identity applications and algebraic manipulations.
- Not applying identities repeatedly: Sometimes one application isn't enough; you might need to apply half-angle identities multiple times until all powers are gone.
- Forgetting the constant of integration ($+C$): This is a universal integration mistake!

5. Now Try It

Take 15 minutes to solve the following integral: $\int x^2 \sin x \, dx$. This will require Integration by Parts twice. What success looks like is arriving at the correct antiderivative, remembering the constant of integration, and clearly showing your steps for each application of IBP.