Integrals Involving Trig Functions for MATH 141
Exam Relevance for MATH 141
Trig integrals (sin^m cos^n) appear on MATH 141 exams. Know strategies for odd/even powers.
This skill appears on:
Understanding Trigonometric Integrals
When you encounter integrals like $\int \sin^4 x \, dx$ or $\int \sec^3 x \tan x \, dx$, basic substitution often isn't enough. These integrals require special techniques based on trig identities and strategic manipulation.
The good news: once you learn a few patterns, most trig integrals follow predictable strategies. This lesson covers the main cases you'll see in Calculus.
Key Identities to Know
Before diving into techniques, make sure you know these identities:
Pythagorean Identities:
- $\sin^2 x + \cos^2 x = 1$
- $\tan^2 x + 1 = \sec^2 x$
- $1 + \cot^2 x = \csc^2 x$
Power-Reducing (Half-Angle) Identities:
- $\sin^2 x = \frac{1 - \cos 2x}{2}$
- $\cos^2 x = \frac{1 + \cos 2x}{2}$
Double Angle Identity:
- $\sin x \cos x = \frac{1}{2}\sin 2x$
Strategy Overview
| Integral Type | Strategy |
|---|---|
| $\int \sin^m x \cos^n x \, dx$ (odd power) | Save one, convert rest with $\sin^2 + \cos^2 = 1$ |
| $\int \sin^m x \cos^n x \, dx$ (both even) | Use power-reducing identities |
| $\int \tan^m x \sec^n x \, dx$ | Depends on powers (see below) |
| $\int \sec x \, dx$ or $\int \csc x \, dx$ | Memorize or use special trick |
Problem: Find $\int \sin^3 x \cos^2 x \, dx$
Step 1: Identify the strategy
We have an odd power of sine (3).
Strategy: Save one $\sin x$ for $du$, convert the rest to cosines.
Step 2: Rewrite and substitute
$$\int \sin^3 x \cos^2 x \, dx$$
$$= \int \sin^2 x \cos^2 x \cdot \sin x \, dx$$
Use $\sin^2 x = 1 - \cos^2 x$:
$$= \int (1 - \cos^2 x) \cos^2 x \cdot \sin x \, dx$$
Step 3: U-substitution
Let $u = \cos x$, so $du = -\sin x \, dx$
$$= -\int (1 - u^2) u^2 \, du$$
$$= -\int (u^2 - u^4) \, du$$
Step 4: Integrate and back-substitute
$$= -\left( \frac{u^3}{3} - \frac{u^5}{5} \right) + C$$
$$= -\frac{\cos^3 x}{3} + \frac{\cos^5 x}{5} + C$$
$$= \boxed{\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C}$$
Problem: Find $\int \sin^2 x \, dx$
Step 1: Apply the power-reducing identity
Since both powers are even (we have $\sin^2 x$ and implicitly $\cos^0 x = 1$), use:
$$\sin^2 x = \frac{1 - \cos 2x}{2}$$
Step 2: Substitute and integrate
$$\int \sin^2 x \, dx = \int \frac{1 - \cos 2x}{2} \, dx$$
$$= \frac{1}{2} \int (1 - \cos 2x) \, dx$$
$$= \frac{1}{2} \left( x - \frac{\sin 2x}{2} \right) + C$$
$$= \boxed{\frac{x}{2} - \frac{\sin 2x}{4} + C}$$
Problem: Find $\int \tan^3 x \sec^2 x \, dx$
Step 1: Identify the strategy
We have $\sec^2 x$, which is the derivative of $\tan x$!
This suggests: let $u = \tan x$.
Step 2: Substitute
Let $u = \tan x$, so $du = \sec^2 x \, dx$
$$\int \tan^3 x \sec^2 x \, dx = \int u^3 \, du$$
Step 3: Integrate and back-substitute
$$= \frac{u^4}{4} + C$$
$$= \boxed{\frac{\tan^4 x}{4} + C}$$
Problem: Find $\int \tan^3 x \sec x \, dx$
Step 1: Identify the strategy
Odd power of tangent with at least one $\sec x$.
Strategy: Save $\sec x \tan x$ for $du$, convert rest to secants.
Step 2: Rewrite
$$\int \tan^3 x \sec x \, dx$$
$$= \int \tan^2 x \cdot \sec x \tan x \, dx$$
Use $\tan^2 x = \sec^2 x - 1$:
$$= \int (\sec^2 x - 1) \sec x \tan x \, dx$$
Step 3: Substitute
Let $u = \sec x$, so $du = \sec x \tan x \, dx$
$$= \int (u^2 - 1) \, du$$
Step 4: Integrate and back-substitute
$$= \frac{u^3}{3} - u + C$$
$$= \boxed{\frac{\sec^3 x}{3} - \sec x + C}$$
Problem: Find $\int \sec x \, dx$
Step 1: Use the classic trick
Multiply by $\frac{\sec x + \tan x}{\sec x + \tan x}$:
$$\int \sec x \, dx$$
$$= \int \sec x \cdot \frac{\sec x + \tan x}{\sec x + \tan x} \, dx$$
$$= \int \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} \, dx$$
Step 2: Recognize the form
The numerator is the derivative of the denominator!
Let $u = \sec x + \tan x$
Then $du = (\sec x \tan x + \sec^2 x) \, dx$
Step 3: Integrate
$$= \int \frac{du}{u} = \ln|u| + C$$
$$= \boxed{\ln|\sec x + \tan x| + C}$$
💡 Tip: Most students just memorize $\int \sec x \, dx = \ln|\sec x + \tan x| + C$. The derivation is clever but rarely needed on exams.
Integrals That Give Logarithms
Sometimes a trig integral simplifies to $\int \frac{1}{u} \, du$, which gives $\ln|u|$. This happens when the numerator is the derivative of the denominator (or can be made into one).
The key pattern: if you see a fraction where the top is (or contains) the derivative of the bottom, think logarithm.
Problem: Find $\int \tan x \, dx$
Step 1: Rewrite in terms of sine and cosine
$$\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx$$
Step 2: Recognize the pattern
The numerator $\sin x$ is almost the derivative of $\cos x$.
(Actually, $\frac{d}{dx}[\cos x] = -\sin x$, so we're off by a negative.)
Step 3: Substitute
Let $u = \cos x$, so $du = -\sin x \, dx$
$$= \int \frac{-du}{u} = -\ln|u| + C$$
Step 4: Back-substitute
$$= -\ln|\cos x| + C$$
$$= \boxed{\ln|\sec x| + C}$$
(Using the log rule: $-\ln|a| = \ln|1/a| = \ln|\sec x|$)
Integrals That Give Inverse Trig Functions
Some trig-related integrals produce arctangent or arcsine. These typically involve expressions like $\frac{1}{1+x^2}$ or $\frac{1}{\sqrt{1-x^2}}$.
Key formulas to recognize:
- $\int \frac{1}{1+x^2} \, dx = \arctan x + C$
- $\int \frac{1}{a^2+x^2} \, dx = \frac{1}{a}\arctan\frac{x}{a} + C$
Problem: Find $\int \frac{1}{9 + x^2} \, dx$
Step 1: Match the formula pattern
Compare to $\int \frac{1}{a^2+x^2} \, dx = \frac{1}{a}\arctan\frac{x}{a} + C$
Here $a^2 = 9$, so $a = 3$.
Step 2: Apply the formula
$$\int \frac{1}{9 + x^2} \, dx = \frac{1}{3}\arctan\frac{x}{3} + C$$
$$= \boxed{\frac{1}{3}\arctan\frac{x}{3} + C}$$
💡 Tip: When you see $\frac{1}{\text{constant} + x^2}$, think arctangent!
Important Results to Memorize
These come up frequently enough that you should know them:
| Integral | Result |
|---|---|
| $\int \sec x \, dx$ | $\ln\|\sec x + \tan x\| + C$ |
| $\int \csc x \, dx$ | $-\ln\|\csc x + \cot x\| + C$ |
| $\int \tan x \, dx$ | $-\ln\|\cos x\| + C$ or $\ln\|\sec x\| + C$ |
| $\int \cot x \, dx$ | $\ln\|\sin x\| + C$ |
Common Mistakes and Misunderstandings
❌ Mistake: Using power-reducing when there's an odd power
Wrong approach: For $\int \sin^3 x \, dx$, trying to use $\sin^2 x = \frac{1-\cos 2x}{2}$
Why it's wrong: This makes the problem harder, not easier. You'd get $\sin x \cdot \frac{1-\cos 2x}{2}$, which is messier.
Correct: Use the odd-power strategy. Save one $\sin x$, convert the rest:
$$\int \sin^3 x \, dx = \int (1-\cos^2 x)\sin x \, dx$$
Then substitute $u = \cos x$.
❌ Mistake: Forgetting which identity to use for sec/tan
Confused: "Do I use $\tan^2 x = \sec^2 x - 1$ or $\sec^2 x = \tan^2 x + 1$?"
Clarification: They're the same identity rearranged! Use whichever form helps you:
- Converting $\tan^2$ to $\sec^2$: use $\tan^2 x = \sec^2 x - 1$
- Converting $\sec^2$ to $\tan^2$: use $\sec^2 x = \tan^2 x + 1$
❌ Mistake: Wrong substitution choice for sec/tan integrals
Wrong: For $\int \sec^4 x \tan x \, dx$, letting $u = \tan x$
Why it's wrong: $du = \sec^2 x \, dx$, but we only have $\sec^4 x$, not a clean match.
Correct: Let $u = \sec x$, so $du = \sec x \tan x \, dx$
$$\int \sec^4 x \tan x \, dx = \int \sec^3 x \cdot \sec x \tan x \, dx = \int u^3 \, du$$
Power-Reducing Identity (Sine)
Converts sine squared into a form without powers. Use when both sine and cosine have even powers.
Variables:
- $\sin^2 x$:
- sine squared, which we want to simplify
- $\cos 2x$:
- cosine of double the angle
Power-Reducing Identity (Cosine)
Converts cosine squared into a form without powers. Use when both sine and cosine have even powers.
Variables:
- $\cos^2 x$:
- cosine squared, which we want to simplify
- $\cos 2x$:
- cosine of double the angle
Pythagorean Identity (Sine/Cosine)
The fundamental trig identity. Rearrange to convert between sine and cosine: sin²x = 1 - cos²x or cos²x = 1 - sin²x.
Variables:
- $\sin^2 x$:
- sine squared
- $\cos^2 x$:
- cosine squared
Pythagorean Identity (Tangent/Secant)
Use to convert between tangent and secant. Rearranged: tan²x = sec²x - 1.
Variables:
- $\tan^2 x$:
- tangent squared
- $\sec^2 x$:
- secant squared
Integral of Secant
A result worth memorizing. Derived using a clever multiplication trick.
Variables:
- $\sec x$:
- secant of x
- $\tan x$:
- tangent of x
Integral of Cosecant
Similar to secant integral. Note the negative sign.
Variables:
- $\csc x$:
- cosecant of x
- $\cot x$:
- cotangent of x
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