# Integration of Trigonometric Functions

Integration of Trigonometric functions involves basic simplification techniques. These techniques use different trigonometric identities which can be written in an alternative form that are more amenable to integration.

## Representation

The integration of a function f(x) is given by F(x) and it is represented by:

 ∫f(x)dx = F(x) + C

Here,

R.H.S. of the equation means integral f(x) with respect to x.

F(x) is called anti-derivative or primitive.

f(x) is called the integrand.

dx is called the integrating agent.

C is called constant of integration or arbitrary constant.

x is the variable of integration.

Also, check integral formulas here.

### Integration of Trigonometric Functions Formulas

Below are the list of few formulas for the integration of trigonometric functions:

• ∫sin x dx = -cos x + C
• ∫cos x dx = sin x + C
• ∫tan x dx = ln|sec x| + C
• ∫sec x dx = ln|tan x + sec x| + C
• ∫cosec x dx = ln|cosec x – cot x| + C = ln|tan(x/2)| + C
• ∫cot x dx = ln|sin x| + C
• ∫sec2x dx = tan x + C
• ∫cosec2x dx = -cot x + C
• ∫sec x tan x dx = sec x + C
• ∫cosec x cot x dx = -cosec x + C
• ∫sin kx dx = -(cos kx/k) + C
• ∫cos kx dx = (sin kx/k) + C

To understand this concept let us solve some examples.

## Integration of Trigonometric Functions Examples

Example 1:

 Question- Integrate 2cos2x with respect to x. Solution- To integrate the given trigonometric functions we will use the trigonometric identity – $$\begin{array}{l}\cos^2 x = \left ( \frac{1 + \cos 2x}{2} \right )\end{array}$$ Form this identity $$\begin{array}{l}2 \cos^2 x = 1 + \cos 2x\end{array}$$ Substituting the above value in the given integrand, we have $$\begin{array}{l}\int 2 \cos^{2}x dx = \int (1+ \cos 2x). dx\end{array}$$ — (1) According to the properties of integration, the integral of sum of two functions is equal to the sum of integrals of the given functions, i.e., $$\begin{array}{l}\int [f(x)+g(x)]dx = \int f(x).dx + \int g(x).dx\end{array}$$ Therefore equation 1 can be rewritten as: $$\begin{array}{l}\int (1 + cos 2x) dx = \int 1 dx + \int cos2xdx\end{array}$$ = $$\begin{array}{l}x + \frac{\sin 2x}{2} + C\end{array}$$ This gives us the required integration of the given function.

Example 2:

 Question- Integrate sin 4x cos 3x with respect to x. Solution- To integrate the trigonometric function we will use the trigonometric identity: $$\begin{array}{l}\sin x \cos y = \frac{1}{2} [\sin (x+y) + \sin (x-y)]\end{array}$$ Form this identity$$\begin{array}{l}\sin 4x \cos 3x = \frac{1}{2} (\sin 7x + \sin x)\end{array}$$ Therefore, $$\begin{array}{l}\int (\sin4x \cos3x)dx = \int \frac{1}{2} (\sin7x + \sin x)dx\end{array}$$ From the above equation we have: $$\begin{array}{l}\int \frac{1}{2}(\sin7x + \sin x)dx = \frac{1}{2} \int (\sin7x + \sin x)dx\end{array}$$ …………(ii) According to the properties of integration, the integral of sum of two functions is equal to the sum of integrals of the given functions, i.e., $$\begin{array}{l}\int [f(x)+g(x)]dx = \int f(x).dx + \int g(x).dx\end{array}$$ Therefore equation 2 can be rewritten as: $$\begin{array}{l}\frac{1}{2} \int (\sin7x ) + \frac{1}{2}\int (\sin x)dx\end{array}$$ = $$\begin{array}{l}\frac{-\cos 7x}{14} + \frac{-\cos x}{2} + C\end{array}$$ This gives us the required integration of the given function.

Example 3:

 Question- Integrate $$\begin{array}{l}\sin ^{2}x. \cos ^{2}x\end{array}$$. Solution- Before integration let us use few trigonometric relations in order to simplify the integrand. We know, $$\begin{array}{l}2 \sin x \cos x = \sin 2x\end{array}$$ $$\begin{array}{l}\sin x . \cos x = \frac{\sin 2x}{2}\end{array}$$ Substituting the value in the given integrand, we have $$\begin{array}{l}\int \sin ^{2}x. \cos ^{2}x \; dx = \int (\sin x . \cos x)^{2} dx = \int \left ( \frac{\sin 2x}{2} \right )^{2}\end{array}$$ $$\begin{array}{l}= \frac{1}{4} \int \sin^{2} 2x\end{array}$$ ………(i) Also we know $$\begin{array}{l}\sin^2 x = \frac{1 – \cos 2x }{2}\end{array}$$ Substituting the above value in equation (i), we have $$\begin{array}{l}\frac{1}{4} \int \sin^{2} 2x = \frac{1}{4} \int \frac{1- \cos 4x}{2}\end{array}$$ $$\begin{array}{l}= \int \frac{1}{8} dx – \int \frac{\cos 4x}{8} \; dx\end{array}$$ $$\begin{array}{l}= \frac{1}{8} x + C_{1} – \frac{\sin 4x}{32} + C_{2}\end{array}$$ $$\begin{array}{l}= \frac{1}{8} x – \frac{\sin 4x}{32} + C\end{array}$$

### Integration of Trigonometric Functions Questions

Try solving the following practical problems on integration of trigonometric functions.

1. Find the integral of (cos x + sin x).
2. Evaluate: ∫(1 – cos x)/sin2x dx
3. Find the integral of sin2x, i.e. ∫sin2x dx.

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