Indefinite Integral of Trigonometric Functions
Smaridasa Mathematics – After studying "Indefinite Integral of Algebraic Functions", we will continue with integral material related to the Indefinite Integral of Trigonometric Functions. The properties of indefinite integrals also apply to trigonometric integrals. For ease, please read the material on "Derivative of Trigonometric Functions" first, because integration is the inverse of differentiation.
Basic Trigonometric Integral Formulas
Based on the definition of the integral: \( \int f(x) \, dx = F(x) + c \), where \( F'(x) = f(x) \):
| Derivative Rule | Corresponding Integral Formula |
|---|---|
| \( \frac{d}{dx} (\sin x) = \cos x \) | \( \displaystyle \int \cos x \, dx = \sin x + c \) |
| \( \frac{d}{dx} (\cos x) = -\sin x \) | \( \displaystyle \int \sin x \, dx = -\cos x + c \) |
| \( \frac{d}{dx} (\tan x) = \sec^2 x \) | \( \displaystyle \int \sec^2 x \, dx = \tan x + c \) |
| \( \frac{d}{dx} (\cot x) = -\csc^2 x \) | \( \displaystyle \int \csc^2 x \, dx = -\cot x + c \) |
| \( \frac{d}{dx} (\sec x) = \sec x \tan x \) | \( \displaystyle \int \sec x \tan x \, dx = \sec x + c \) |
| \( \frac{d}{dx} (\csc x) = -\csc x \cot x \) | \( \displaystyle \int \csc x \cot x \, dx = -\csc x + c \) |
Recall: \( \sec x = \frac{1}{\cos x} \), \( \csc x = \frac{1}{\sin x} \), \( \tan x = \frac{\sin x}{\cos x} \), \( \cot x = \frac{\cos x}{\sin x} \).
Example Set 1: Basic Trigonometric Integrals
1a. \( \displaystyle \int 2\sin x \, dx \)
\[ \int 2\sin x \, dx = 2 \int \sin x \, dx = 2(-\cos x) + c = -2\cos x + c \]
1b. \( \displaystyle \int (\cos x - \sin x) \, dx \)
\[ \int (\cos x - \sin x) \, dx = \int \cos x \, dx - \int \sin x \, dx = \sin x - (-\cos x) + c = \sin x + \cos x + c \]
1c. \( \displaystyle \int \frac{\sin x + \csc x}{\tan x} \, dx \)
Simplify first:
\[ \frac{\sin x + \csc x}{\tan x} = (\sin x + \csc x) \cdot \frac{\cos x}{\sin x} = \cos x + \csc x \cot x \]
\[ \int (\cos x + \csc x \cot x) \, dx = \sin x - \csc x + c \]
1d. \( \displaystyle \int \frac{\tan x + \cot x}{\sin 2x} \, dx \)
Use \( \sin 2x = 2 \sin x \cos x \):
\[ \frac{\tan x + \cot x}{\sin 2x} = \frac{\frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}}{2 \sin x \cos x} = \frac{1}{2\cos^2 x} + \frac{1}{2\sin^2 x} = \frac{1}{2}(\sec^2 x + \csc^2 x) \]
\[ \int \frac{1}{2}(\sec^2 x + \csc^2 x) \, dx = \frac{1}{2} (\tan x - \cot x) + c \]
Integrals with Linear Arguments: \( \sin(ax+b) \), \( \cos(ax+b) \), etc.
General formulas for trigonometric functions with linear arguments:
| Integral | Result |
|---|---|
| \( \displaystyle \int \sin(ax+b) \, dx \) | \( \displaystyle -\frac{1}{a} \cos(ax+b) + c \) |
| \( \displaystyle \int \cos(ax+b) \, dx \) | \( \displaystyle \frac{1}{a} \sin(ax+b) + c \) |
| \( \displaystyle \int \sec^2(ax+b) \, dx \) | \( \displaystyle \frac{1}{a} \tan(ax+b) + c \) |
| \( \displaystyle \int \csc^2(ax+b) \, dx \) | \( \displaystyle -\frac{1}{a} \cot(ax+b) + c \) |
| \( \displaystyle \int \sec(ax+b)\tan(ax+b) \, dx \) | \( \displaystyle \frac{1}{a} \sec(ax+b) + c \) |
| \( \displaystyle \int \csc(ax+b)\cot(ax+b) \, dx \) | \( \displaystyle -\frac{1}{a} \csc(ax+b) + c \) |
Example Set 2: Integrals with Linear Arguments
2a. \( \displaystyle \int \sin(2x + 3) \, dx \)
\[ \int \sin(2x + 3) \, dx = -\frac{1}{2} \cos(2x + 3) + c \]
2b. \( \displaystyle \int 6\cos(1 - 3x) \, dx \)
\[ \int 6\cos(1 - 3x) \, dx = 6 \cdot \frac{1}{-3} \sin(1 - 3x) + c = -2 \sin(1 - 3x) + c \]
2c. \( \displaystyle \int \sec^2(4x) \, dx \)
\[ \int \sec^2(4x) \, dx = \frac{1}{4} \tan(4x) + c \]
2d. \( \displaystyle \int [\csc^2(-2x + 1) + \sin(2x)] \, dx \)
Integrate each term separately:
\[ \int \csc^2(-2x + 1) \, dx = -\frac{1}{-2} \cot(-2x + 1) + c_1 = \frac{1}{2} \cot(-2x + 1) + c_1 \]
\[ \int \sin(2x) \, dx = -\frac{1}{2} \cos(2x) + c_2 \]
\[ \text{Total} = \frac{1}{2} \cot(-2x + 1) - \frac{1}{2} \cos(2x) + c \]
2e. \( \displaystyle \int \sec(x - 1) \tan(x - 1) \, dx \)
\[ \int \sec(x - 1) \tan(x - 1) \, dx = \sec(x - 1) + c \]
2f. \( \displaystyle \int \csc(5x - 3) \cot(5x - 3) \, dx \)
\[ \int \csc(5x - 3) \cot(5x - 3) \, dx = -\frac{1}{5} \csc(5x - 3) + c \]
Product-to-Sum Formulas for Integration
Useful trigonometric identities for integrating products:
\[ \sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)] \]
\[ \cos A \sin B = \frac{1}{2} [\sin(A + B) - \sin(A - B)] \]
\[ \cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)] \]
\[ \sin A \sin B = -\frac{1}{2} [\cos(A + B) - \cos(A - B)] \]
Also useful: \( \cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta) \), \( \sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta) \).
Example Set 3: Using Product-to-Sum Formulas
3a. \( \displaystyle \int \sin 5x \cos 2x \, dx \)
Use \( \sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)] \):
\[ \int \sin 5x \cos 2x \, dx = \frac{1}{2} \int [\sin(7x) + \sin(3x)] \, dx \]
\[ = \frac{1}{2} \left[ -\frac{1}{7}\cos(7x) - \frac{1}{3}\cos(3x) \right] + c \]
3b. \( \displaystyle \int 4\cos 7x \sin 4x \, dx \)
Use \( \cos A \sin B = \frac{1}{2}[\sin(A+B) - \sin(A-B)] \):
\[ \int 4\cos 7x \sin 4x \, dx = 2 \int [\sin(11x) - \sin(3x)] \, dx \]
\[ = 2 \left[ -\frac{1}{11}\cos(11x) + \frac{1}{3}\cos(3x) \right] + c \]
3c. \( \displaystyle \int 3\cos(3x-1) \cos(2x+2) \, dx \)
Use \( \cos A \cos B = \frac{1}{2}[\cos(A+B) + \cos(A-B)] \):
\[ \int 3\cos(3x-1) \cos(2x+2) \, dx = \frac{3}{2} \int [\cos(5x+1) + \cos(x-3)] \, dx \]
\[ = \frac{3}{2} \left[ \frac{1}{5}\sin(5x+1) + \sin(x-3) \right] + c \]
3d. \( \displaystyle \int \cos^2 3x \, dx \)
Use \( \cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta) \):
\[ \int \cos^2 3x \, dx = \frac{1}{2} \int (1 + \cos 6x) \, dx = \frac{1}{2} \left( x + \frac{1}{6}\sin 6x \right) + c \]
3e. \( \displaystyle \int \sin^4 5x \, dx \)
First rewrite \( \sin^4 5x = (\sin^2 5x)^2 \), then use \( \sin^2 \theta = \frac{1}{2}(1 - \cos 2\theta) \):
\[ \sin^4 5x = \left[ \frac{1}{2}(1 - \cos 10x) \right]^2 = \frac{1}{4}(1 - 2\cos 10x + \cos^2 10x) \]
Use \( \cos^2 10x = \frac{1}{2}(1 + \cos 20x) \):
\[ \sin^4 5x = \frac{1}{4}\left[ 1 - 2\cos 10x + \frac{1}{2}(1 + \cos 20x) \right] = \frac{1}{4}\left( \frac{3}{2} - 2\cos 10x + \frac{1}{2}\cos 20x \right) \]
Now integrate term by term:
\[ \int \sin^4 5x \, dx = \frac{1}{4} \left[ \frac{3}{2}x - \frac{2}{10}\sin 10x + \frac{1}{2} \cdot \frac{1}{20}\sin 20x \right] + c \]
\[ = \frac{3}{8}x - \frac{1}{20}\sin 10x + \frac{1}{160}\sin 20x + c \]