Integrating Trigonometric Rational Functions

What You'll Learn

• Definition and general methods for trigonometric rational functions
• Weierstrass substitution (t = tan(x/2)): principles and applications
• 8 special solution techniques using trigonometric identities
• Strategic case-by-case analysis
• 6 comprehensive examples, from fundamental to advanced

1. Overview of Trigonometric Rational Function Integration

Definition 1.1: Trigonometric Rational Functions

Integrals of the form $\int R(\sin x, \cos x)\,dx$ are called trigonometric rational function integrals, where $R(u, v)$ is a rational function in $u, v$ .

Examples: $\int \frac{\sin x}{1 + \cos x}\,dx$ , $\int \frac{1}{3 - 5\cos x}\,dx$ , etc.

Solution Strategy

Trigonometric rational function integrals typically follow three solution paths:

Weierstrass Substitution: Universal method that always reduces to rational function integration
Trigonometric Identity Transformations: Use sum-to-product, double-angle formulas, etc. to simplify
Special Techniques: Choose clever methods based on the integrand's characteristics

2. Weierstrass Substitution

Theorem 2.1: Weierstrass Substitution Formulas

Let $t = \tan\frac{x}{2}$ , then:

\sin x = \frac{2t}{1+t^2}

\cos x = \frac{1-t^2}{1+t^2}

dx = \frac{2}{1+t^2}\,dt

This transforms the trigonometric rational function integral into a rational function integral in $t$ .

Usage Precautions

While "universal," this substitution can lead to computational complexity
First consider whether trigonometric identities can simplify the problem
When the integrand has special symmetry, simpler methods often exist
Pay attention to domain changes after substitution

3. Classification of Special Solution Methods

Case 1: Numerator is derivative of denominator

If $R(\sin x, \cos x)$ can be written as:

\frac{f'(\sin x, \cos x)}{f(\sin x, \cos x)}

Then the result is directly $\ln|f(\sin x, \cos x)| + C$

Case 2: Power-reduction formulas

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

Use reduction formulas or $\sin^2 x = \frac{1-\cos 2x}{2}$

Case 3: Sum-to-product identities

For $\sin ax \cdot \sin bx$ type:

\sin ax \sin bx = \frac{1}{2}[\cos(a-b)x - \cos(a+b)x]

Case 4: Homogenization

For $\frac{a\sin x + b\cos x}{c\sin x + d\cos x}$ :

Set integrand = $\lambda(c\sin x + d\cos x)' + \mu$

4. Detailed Example Solutions

Example 4.1: Standard Application of Weierstrass Substitution

Problem: Find $\int \frac{1}{3-5\cos x}\,dx$

Solution:

Step 1: Apply Weierstrass substitution

Let $t = \tan\frac{x}{2}$ , then:

\cos x = \frac{1-t^2}{1+t^2}, \quad dx = \frac{2}{1+t^2}\,dt

Step 2: Substitute and simplify

\begin{aligned} I &= \int \frac{1}{3-5\cdot\frac{1-t^2}{1+t^2}} \cdot \frac{2}{1+t^2}\,dt \\ &= \int \frac{2}{3(1+t^2) - 5(1-t^2)}\,dt \\ &= \int \frac{2}{3 + 3t^2 - 5 + 5t^2}\,dt \\ &= \int \frac{2}{8t^2 - 2}\,dt = \int \frac{1}{4t^2 - 1}\,dt \end{aligned}

Step 3: Partial fraction decomposition

\frac{1}{4t^2-1} = \frac{1}{(2t-1)(2t+1)} = \frac{1}{2}\left(\frac{1}{2t-1} - \frac{1}{2t+1}\right)

Step 4: Integrate and back-substitute

\begin{aligned} I &= \frac{1}{2}\int\left(\frac{1}{2t-1} - \frac{1}{2t+1}\right)dt \\ &= \frac{1}{4}\ln|2t-1| - \frac{1}{4}\ln|2t+1| + C \\ &= \frac{1}{4}\ln\left|\frac{2t-1}{2t+1}\right| + C \\ &= \frac{1}{4}\ln\left|\frac{2\tan\frac{x}{2}-1}{2\tan\frac{x}{2}+1}\right| + C \end{aligned}

Answer

I = \frac{1}{4}\ln\left|\frac{2\tan\frac{x}{2}-1}{2\tan\frac{x}{2}+1}\right| + C

Example 4.2: Simplification via Trigonometric Identities

Problem: Find $\int \frac{1}{1+\sin x+\cos x}\,dx$

Solution:

Method 1: Trigonometric Identities (Recommended)

Step 1: Apply half-angle formulas

Use $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ and $\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$ :

\begin{aligned} 1 + \sin x + \cos x &= \sin^2\frac{x}{2} + \cos^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} + \cos^2\frac{x}{2} - \sin^2\frac{x}{2} \\ &= 2\cos^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2} \\ &= 2\cos\frac{x}{2}\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right) \end{aligned}

Step 2: Substitute into integral

I = \int \frac{1}{2\cos\frac{x}{2}\left(\cos\frac{x}{2} + \sin\frac{x}{2}\right)}\,dx

Let $u = \tan\frac{x}{2}$ , then $du = \frac{1}{2}\sec^2\frac{x}{2}\,dx$ :

I = \int \frac{1}{u+1}\,du = \ln|u+1| + C = \ln\left|\tan\frac{x}{2}+1\right| + C

Method 2: Weierstrass Substitution

Let $t = \tan\frac{x}{2}$ yields the same result, but with more tedious calculations.

Answer

I = \ln\left|1 + \tan\frac{x}{2}\right| + C

Example 4.3: Matching Derivative in Numerator

Problem: Find $\int \frac{1}{1-\cos x}\,dx$

Solution:

Step 1: Apply half-angle formula

1 - \cos x = 2\sin^2\frac{x}{2}

Step 2: Substitute into integral

\begin{aligned} I &= \int \frac{1}{2\sin^2\frac{x}{2}}\,dx = \frac{1}{2}\int \csc^2\frac{x}{2}\,dx \\ &= \frac{1}{2} \cdot (-2\cot\frac{x}{2}) + C = -\cot\frac{x}{2} + C \end{aligned}

Verification

Differentiate: $\frac{d}{dx}(-\cot\frac{x}{2}) = \frac{1}{2}\csc^2\frac{x}{2} = \frac{1}{2\sin^2\frac{x}{2}} = \frac{1}{1-\cos x}$ ✓

Answer

I = -\cot\frac{x}{2} + C

Example 4.4: Homogenization Method

Problem: Find $\int \frac{\cos x}{\sin x + \cos x}\,dx$

Solution:

Approach: Undetermined Coefficients

Express the numerator as a linear combination of the denominator's derivative plus the denominator itself:

\cos x = \lambda(\sin x + \cos x)' + \mu(\sin x + \cos x)

Step 1: Find undetermined coefficients

Since $(\sin x + \cos x)' = \cos x - \sin x$ , we have:

\cos x = \lambda(\cos x - \sin x) + \mu(\sin x + \cos x)

Compare coefficients:

\begin{cases} \lambda + \mu = 1 \quad (\text{coeff. of } \cos x) \\ -\lambda + \mu = 0 \quad (\text{coeff. of } \sin x) \end{cases}

Solving: $\lambda = \frac{1}{2}, \mu = \frac{1}{2}$

Step 2: Decompose the integral

\begin{aligned} I &= \int \frac{\frac{1}{2}(\cos x - \sin x) + \frac{1}{2}(\sin x + \cos x)}{\sin x + \cos x}\,dx \\ &= \frac{1}{2}\int \frac{\cos x - \sin x}{\sin x + \cos x}\,dx + \frac{1}{2}\int 1\,dx \\ &= \frac{1}{2}\ln|\sin x + \cos x| + \frac{x}{2} + C \end{aligned}

Answer

I = \frac{1}{2}\ln|\sin x + \cos x| + \frac{x}{2} + C

Example 4.5: Power Reduction for High Powers

Problem: Find $\int \sin^3 x \cos^3 x\,dx$

Solution:

Step 1: Extract a factor

\begin{aligned} I &= \int \sin^3 x \cos^3 x\,dx = \int \sin^2 x \cos^2 x \cdot \sin x \cos x\,dx \\ &= \int (\sin x \cos x)^2 \cdot \sin x \cos x\,dx \end{aligned}

Step 2: Apply double-angle formula

Use $\sin 2x = 2\sin x \cos x$ :

\begin{aligned} I &= \int \left(\frac{\sin 2x}{2}\right)^2 \cdot \frac{\sin 2x}{2}\,dx \\ &= \frac{1}{8}\int \sin^3 2x\,dx = \frac{1}{8}\int \sin^2 2x \cdot \sin 2x\,dx \end{aligned}

Step 3: Use $\sin^2 2x = 1 - \cos^2 2x$

\begin{aligned} I &= \frac{1}{8}\int (1 - \cos^2 2x)\sin 2x\,dx \\ &= \frac{1}{8}\int (1 - \cos^2 2x)\,d(-\frac{\cos 2x}{2}) \\ &= -\frac{1}{16}\int (1 - \cos^2 2x)\,d(\cos 2x) \end{aligned}

Let $u = \cos 2x$ :

I = -\frac{1}{16}\int (1-u^2)\,du = -\frac{1}{16}\left(u - \frac{u^3}{3}\right) + C

Step 4: Back-substitute

I = -\frac{1}{16}\cos 2x + \frac{1}{48}\cos^3 2x + C

Answer

I = -\frac{1}{16}\cos 2x + \frac{1}{48}\cos^3 2x + C

Example 4.6: Complex Comprehensive Problem

Problem: Find $\int \frac{5-4\cos x}{(2+\cos x)^2\sin x}\,dx$

Solution:

Step 1: Attempt numerator adjustment

Notice the denominator is $(2+\cos x)^2\sin x$ . We could try expressing the numerator in terms of the denominator's derivative.

Set:

5 - 4\cos x = A\frac{d}{dx}[(2+\cos x)^2] + B(2+\cos x)^2\cdot\frac{d}{dx}(\sin x)

Step 2: Weierstrass substitution (simpler approach)

Let $t = \tan\frac{x}{2}$ , then:

\begin{aligned} \sin x &= \frac{2t}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2} \\ 5 - 4\cos x &= 5 - 4\cdot\frac{1-t^2}{1+t^2} = \frac{5(1+t^2) - 4(1-t^2)}{1+t^2} = \frac{1+9t^2}{1+t^2} \\ 2 + \cos x &= 2 + \frac{1-t^2}{1+t^2} = \frac{3+t^2}{1+t^2} \end{aligned}

Step 3: Substitute and simplify

\begin{aligned} I &= \int \frac{\frac{1+9t^2}{1+t^2}}{\left(\frac{3+t^2}{1+t^2}\right)^2 \cdot \frac{2t}{1+t^2}} \cdot \frac{2}{1+t^2}\,dt \\ &= \int \frac{(1+9t^2) \cdot 2 \cdot (1+t^2)}{(3+t^2)^2 \cdot 2t \cdot (1+t^2)}\,dt \\ &= \int \frac{1+9t^2}{t(3+t^2)^2}\,dt \end{aligned}

This integral requires partial fraction decomposition, which is fairly involved.

Note

This is a computationally intensive problem requiring patient partial fraction work. In actual exams, such problems typically come with additional hints or simplifications.

5. Method Selection Decision Tree

1

Observe special structure

Can half-angle formulas be applied directly? e.g., $\frac{1}{1\pm\cos x}$
Is the numerator the derivative of the denominator?
Can homogenization work? e.g., $\frac{a\sin x + b\cos x}{c\sin x + d\cos x}$

2

Try trigonometric identities

Sum-to-product, double-angle, power-reduction formulas
Can common factors be extracted to simplify?

3

Use Weierstrass substitution

When the first two approaches fail, the Weierstrass substitution $t = \tan\frac{x}{2}$ will always work

6. Common Pitfalls

Pitfall 1

Domain issues with Weierstrass substitution

$t = \tan\frac{x}{2}$ is undefined at $x = (2k+1)\pi$ ; pay attention to integration intervals

Pitfall 2

Incorrect trigonometric identity signs

Note the signs in $\sin^2 x = \frac{1-\cos 2x}{2}$ and $\cos^2 x = \frac{1+\cos 2x}{2}$

Pitfall 3

Signs during back-substitution

When converting from $t$ back to $x$ , watch for absolute value signs and positive/negative signs

Pitfall 4

Integration constant

Always remember to include the constant of integration $C$

Integration of Trigonometric Rational Functions

1. Overview of Trigonometric Rational Function Integration

Solution Strategy

2. Weierstrass Substitution

Usage Precautions

3. Classification of Special Solution Methods

4. Detailed Example Solutions

5. Method Selection Decision Tree

6. Common Pitfalls