3.2 Trigonometric Integrals

Trigonometric Integrals

Trigonometric integrals involve integrating products and powers of trig functions like sine, cosine, tangent, and secant. The core challenge is that you can't just use the power rule on something like $\sin^5 x \cos^2 x$. Instead, you need specific strategies depending on whether the powers are odd or even, and which trig functions are involved.

This section covers those strategies: Pythagorean identity tricks, half-angle formulas, product-to-sum conversions, and reduction formulas.

Integration of Sine and Cosine Products

The general form here is $\int \sin^m x \cos^n x \, dx$. Your approach depends entirely on whether $m$ and $n$ are odd or even.

When at least one power is odd

If either $m$ or $n$ is odd, you can peel off one factor of that function to pair with $dx$, then convert everything remaining using the Pythagorean identity.

Steps (say $n$ is odd):

1. Peel off one $\cos x$ and set it aside with $dx$

2. Rewrite the remaining $\cos^{n-1} x$ (now an even power) using $\cos^2 x = 1 - \sin^2 x$

3. Substitute $u = \sin x$, so $du = \cos x \, dx$

4. You now have a polynomial in $u$ that you can integrate with the power rule

The same logic works if $m$ is odd: peel off one $\sin x$, convert the rest using $\sin^2 x = 1 - \cos^2 x$, and substitute $u = \cos x$.

Example: For $\int \sin^2 x \cos^3 x \, dx$, the cosine power (3) is odd. Peel off one $\cos x$, rewrite $\cos^2 x = 1 - \sin^2 x$, and substitute $u = \sin x$.

When both powers are even

If both $m$ and $n$ are even, the Pythagorean identity won't help because it just trades one even power for another. Instead, use the half-angle formulas to reduce the powers:

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

Substitute these in, multiply out, and integrate. You may need to apply the half-angle formulas more than once if higher even powers remain.

Example: For $\int \sin^2 x \cos^2 x \, dx$, replace both using half-angle formulas, then expand and simplify before integrating.

Products with different arguments

For integrals like $\int \sin(mx) \cos(nx) \, dx$ where the arguments differ, use product-to-sum formulas:

• $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$
• $\cos A \sin B = \frac{1}{2}[\sin(A+B) - \sin(A-B)]$
• $\sin A \sin B = \frac{1}{2}[\cos(A-B) - \cos(A+B)]$
• $\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]$

These convert the product into a sum of single trig functions, which you can integrate directly.

Integrals with Tangent and Secant

These follow a different set of strategies built around the identity $\tan^2 x = \sec^2 x - 1$.

Powers of tangent: $\int \tan^n x \, dx$

For even powers, factor out $\tan^2 x$ and replace it with $\sec^2 x - 1$. This breaks the integral into a piece with $\sec^2 x$ (which pairs nicely with $u = \tan x$) and a lower-power tangent integral.

For odd powers, the same identity works. Peel off a $\tan^2 x$, replace with $\sec^2 x - 1$, and repeat until you reduce to $\int \tan x \, dx = \ln|\sec x| + C$.

Powers of secant: $\int \sec^n x \, dx$

Even powers are the easier case:

1. Peel off $\sec^2 x$ and set it aside with $dx$
2. Convert the remaining $\sec^{n-2} x$ using $\sec^2 x = 1 + \tan^2 x$
3. Substitute $u = \tan x$, so $du = \sec^2 x \, dx$

Odd powers (for $n \geq 3$) are harder and typically require integration by parts. The standard approach for $\int \sec^3 x \, dx$ is a classic IBP problem worth memorizing:

$\int \sec^3 x \, dx = \frac{1}{2}\sec x \tan x + \frac{1}{2}\ln|\sec x + \tan x| + C$

Products of secant and tangent: $\int \sec^m x \tan^n x \, dx$

The strategy depends on the powers:

• If $n$ is odd: Peel off $\sec x \tan x$ (which is the derivative of $\sec x$), convert remaining $\tan^2 x$ factors to $\sec^2 x - 1$, and substitute $u = \sec x$
• If $m$ is even: Peel off $\sec^2 x$ (the derivative of $\tan x$), convert remaining $\sec^2 x$ factors to $1 + \tan^2 x$, and substitute $u = \tan x$

Reduction Formulas for Trigonometric Integrals

Reduction formulas let you express an integral with power $n$ in terms of the same integral with a lower power. They're derived using integration by parts and are especially useful for high powers.

Sine: $\int \sin^n x \, dx = -\frac{1}{n} \sin^{n-1} x \cos x + \frac{n-1}{n} \int \sin^{n-2} x \, dx$

Cosine: $\int \cos^n x \, dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n} \int \cos^{n-2} x \, dx$

Tangent: $\int \tan^n x \, dx = \frac{1}{n-1} \tan^{n-1} x - \int \tan^{n-2} x \, dx$

Secant: $\int \sec^n x \, dx = \frac{1}{n-1} \sec^{n-2} x \tan x + \frac{n-2}{n-1} \int \sec^{n-2} x \, dx$

Each formula reduces the power by 2 (except tangent, which reduces by 2 as well since you go from $n$ to $n-2$). You apply them repeatedly until you reach a base case you know:

• $\int \sin x \, dx = -\cos x + C$ or $\int \sin^0 x \, dx = x + C$
• $\int \tan x \, dx = \ln|\sec x| + C$ or $\int \tan^0 x \, dx = x + C$
• $\int \sec x \, dx = \ln|\sec x + \tan x| + C$ or $\int \sec^2 x \, dx = \tan x + C$

Quick Strategy Summary