Trigonometric Integrals

By the end of this lesson, you will be able to evaluate integrals of the form $\int\sin^m x\cos^n x,dx$, $\int\tan^m x\sec^n x,dx$, and use power-reducing and Pythagorean identities strategically.

Strategy for $\int\sin^m x\cos^n x,dx$

Case 1: $m$ or $n$ is odd.

Save one factor (e.g., $\sin x,dx$ or $\cos x,dx$) and convert the rest using $\sin^2 + \cos^2 = 1$.

Let $u = \cos x$ (if $m$ is odd) or $u = \sin x$ (if $n$ is odd).

Case 2: Both $m$ and $n$ are even.

Use half-angle formulas: $\sin^2 x = \frac{1-\cos 2x}{2}$, $\cos^2 x = \frac{1+\cos 2x}{2}$.

Example 1 — Odd Exponent

$$\int\sin^3 x\cos^2 x,dx$$

$m = 3$ is odd. Save one $\sin x$:

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

$u = \cos x$, $du = -\sin x,dx$:

$= -\int(1-u^2)u^2,du = -\int(u^2-u^4),du = -\frac{u^3}{3}+\frac{u^5}{5}+C = -\frac{\cos^3 x}{3}+\frac{\cos^5 x}{5}+C$

Example 2 — Both Even

$$\int\sin^2 x\cos^2 x,dx$$

$= \int\frac{1-\cos 2x}{2}\cdot\frac{1+\cos 2x}{2},dx = \frac{1}{4}\int(1-\cos^2 2x),dx$

$= \frac{1}{4}\int\sin^2 2x,dx = \frac{1}{4}\int\frac{1-\cos 4x}{2},dx = \frac{x}{8} - \frac{\sin 4x}{32} + C$

Strategy for $\int\tan^m x\sec^n x,dx$

Case 1: $n$ is even. Save $\sec^2 x$, convert rest using $\sec^2 = 1 + \tan^2$. Let $u = \tan x$.

Case 2: $m$ is odd. Save $\sec x\tan x$, convert rest using $\tan^2 = \sec^2-1$. Let $u = \sec x$.

Example 3 — Tangent-Secant (Even Secant)

$$\int\tan^3 x\sec^4 x,dx$$

$n = 4$ even: save $\sec^2 x$, convert $\sec^2 x = 1 + \tan^2 x$:

$= \int\tan^3 x(1+\tan^2 x)\sec^2 x,dx$

$u = \tan x$, $du = \sec^2 x,dx$:

$= \int u^3(1+u^2),du = \frac{u^4}{4}+\frac{u^6}{6}+C = \frac{\tan^4 x}{4}+\frac{\tan^6 x}{6}+C$

Key Integrals to Memorize

$$\int\sec x,dx = \ln|\sec x+\tan x|+C$$

$$\int\csc x,dx = -\ln|\csc x+\cot x|+C$$

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

Example 4 — $\int\sec^3 x,dx$ (Important)

By integration by parts: $u = \sec x$, $dv = \sec^2 x,dx$:

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

$= \sec x\tan x - \int\sec^3 x,dx + \int\sec x,dx$

Let $I = \int\sec^3 x,dx$:

$I = \sec x\tan x - I + \ln|\sec x+\tan x|$

$2I = \sec x\tan x + \ln|\sec x+\tan x|$

$I = \frac{1}{2}(\sec x\tan x+\ln|\sec x+\tan x|)+C$ ✓