Welcome back to Professor Baker’s Math Class! In this session, we are diving into Section 7.2: Trigonometric Integrals. If you have ever felt intimidated by integrals filled with high powers of sines, cosines, tangents, or secants, this guide is for you. The key to success here isn’t just memorizing formulas; it’s recognizing which strategy to apply based on the exponents.

Part 1: Powers of Sine and Cosine

When dealing with integrals in the form $\int \sin^m x \cos^n x \, dx$, our approach depends entirely on whether the powers are odd or even.

Strategy A: At Least One Odd Power

If either sine or cosine has an odd power, you represent the "easy case." The strategy is to save one factor of the odd-powered function to serve as your $du$ and convert the rest using the Pythagorean identity $\sin^2 x + \cos^2 x = 1$.

  • If cosine is odd ($n = 2k + 1$): Save one factor of $\cos x$. Convert the remaining $\cos^2 x$ terms to $(1 - \sin^2 x)$. Let $u = \sin x$.
  • If sine is odd ($m = 2k + 1$): Save one factor of $\sin x$. Convert the remaining $\sin^2 x$ terms to $(1 - \cos^2 x)$. Let $u = \cos x$.

Example from Class: Consider $\int \sin^5 x \cos^2 x \, dx$.
Since the power of sine (5) is odd, we save one $\sin x$ and convert the remaining $\sin^4 x$:

$$\int \sin^4 x \cos^2 x (\sin x \, dx) = \int (1-\cos^2 x)^2 \cos^2 x \sin x \, dx$$

By setting $u = \cos x$ (and consequently $du = -\sin x \, dx$), this becomes a simple polynomial integral!

Strategy B: Both Powers Are Even

If both sine and cosine have even powers, we cannot use the substitution method immediately. Instead, we must use the Half-Angle Identities to reduce the powers:

  • $\sin^2 x = \frac{1}{2}(1 - \cos 2x)$
  • $\cos^2 x = \frac{1}{2}(1 + \cos 2x)$

Note: These problems can be lengthy, as seen in the class notes for $\int \sin^2 t \cos^4 t \, dt$, often requiring you to apply the identities multiple times.

Part 2: Powers of Tangent and Secant

Integrals involving $\tan x$ and $\sec x$ rely on the identity $1 + \tan^2 x = \sec^2 x$. Here, we look for derived pairs.

Strategy A: Power of Secant is Even

If the power of secant is even ($n$ is even and $n \ge 2$), save a factor of $\sec^2 x$. This works perfectly because $\frac{d}{dx}(\tan x) = \sec^2 x$.

Example: $\int \tan^6 x \sec^4 x \, dx$.
We separate $\sec^2 x$ for our $du$:

$$\int \tan^6 x \sec^2 x (\sec^2 x \, dx) = \int \tan^6 x (1 + \tan^2 x) \sec^2 x \, dx$$

Now, simply let $u = \tan x$.

Strategy B: Power of Tangent is Odd

If the power of tangent is odd ($m$ is odd), save a factor of $\sec x \tan x$. This works because $\frac{d}{dx}(\sec x) = \sec x \tan x$. You will then convert the remaining tangents to secants.

Key Takeaways

  1. Check your Powers: The exponent tells you which identity to use.
  2. Pythagorean Identities are your best friend: Memorize $\sin^2 x + \cos^2 x = 1$ and $1 + \tan^2 x = \sec^2 x$.
  3. Don't panic on Even/Even sines and cosines: Just apply the half-angle formulas patiently.

Happy integrating! Be sure to review the attached slides for the full step-by-step derivation of the more complex examples.