Integration of Trigonometric Functions - Formulas, Solved Examples - GMS - Learning Simply
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# Integration of Trigonometric Functions - Formulas, Solved Examples

Before integration let us use few trigonometric relations in order to simplify the integrand.
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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
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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 – $\begin{array}{l}{\mathrm{cos}}^{2}x=\left(\frac{1+\mathrm{cos}2x}{2}\right)\end{array}$ Form this identity $\begin{array}{l}2{\mathrm{cos}}^{2}x=1+\mathrm{cos}2x\end{array}$ Substituting the above value in the given integrand, we have $\begin{array}{l}\int 2{\mathrm{cos}}^{2}xdx=\int \left(1+\mathrm{cos}2x\right).dx\dots \left(1\right)\end{array}$ 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., $\begin{array}{l}\int \left[f\left(x\right)+g\left(x\right)\right]dx=\int f\left(x\right).dx+\int g\left(x\right).dx\end{array}$ Therefore equation 1 can be rewritten as: $\begin{array}{l}\int \left(1+cos2x\right)dx=\int 1dx+\int cos2xdx\end{array}$ $\begin{array}{l}=x+\frac{\mathrm{sin}2x}{2}+C\end{array}$ 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: $\begin{array}{l}\mathrm{sin}x\mathrm{cos}y=\frac{1}{2}\left[\mathrm{sin}\left(x+y\right)+\mathrm{sin}\left(x-y\right)\right]\end{array}$ $\begin{array}{l}\text{Form this identity}\mathrm{sin}4x\mathrm{cos}3x=\frac{1}{2}\left(\mathrm{sin}7x+\mathrm{sin}x\right)\end{array}$ Therefore, $\begin{array}{l}\int \left(\mathrm{sin}4x\mathrm{cos}3x\right)dx=\int \frac{1}{2}\left(\mathrm{sin}7x+\mathrm{sin}x\right)dx\end{array}$ From the above equation we have: $\begin{array}{l}\int \frac{1}{2}\left(\mathrm{sin}7x+\mathrm{sin}x\right)dx=\frac{1}{2}\int \left(\mathrm{sin}7x+\mathrm{sin}x\right)dx\dots \left(ii\right)\end{array}$ 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., $\begin{array}{l}\int \left[f\left(x\right)+g\left(x\right)\right]dx=\int f\left(x\right).dx+\int g\left(x\right).dx\end{array}$ Therefore equation 2 can be rewritten as: $\begin{array}{l}\frac{1}{2}\int \left(\mathrm{sin}7x\right)+\frac{1}{2}\int \left(\mathrm{sin}x\right)dx\end{array}$ $\begin{array}{l}=\frac{-\mathrm{cos}7x}{14}+\frac{-\mathrm{cos}x}{2}+C\end{array}$ This gives us the required integration of the given function.

Example 3:

 Question- Integrate sin2 x. cos2x. Solution- Before integration let us use few trigonometric relations in order to simplify the integrand. $\begin{array}{l}\mathrm{sin}x.\mathrm{cos}x=\frac{\mathrm{sin}2x}{2}\end{array}$ Substituting the value in the given integrand, we have $\begin{array}{l}\int {\mathrm{sin}}^{2}x.{\mathrm{cos}}^{2}x\phantom{\rule{0.278em}{0ex}}dx=\int \left(\mathrm{sin}x.\mathrm{cos}x{\right)}^{2}dx=\int {\left(\frac{\mathrm{sin}2x}{2}\right)}^{2}\end{array}$ $\begin{array}{l}=\frac{1}{4}\int {\mathrm{sin}}^{2}2x\dots \left(i\end{array}$ Substituting the above value in equation (i), we have $\begin{array}{l}\frac{1}{4}\int {\mathrm{sin}}^{2}2x=\frac{1}{4}\int \frac{1-\mathrm{cos}4x}{2}\end{array}$ $\begin{array}{l}=\int \frac{1}{8}dx–\int \frac{\mathrm{cos}4x}{8}\phantom{\rule{0.278em}{0ex}}dx\end{array}$ $\begin{array}{l}=\frac{1}{8}x+{C}_{1}–\frac{\mathrm{sin}4x}{32}+{C}_{2}\end{array}$ $\begin{array}{l}=\frac{1}{8}x–\frac{\mathrm{sin}4x}{32}+C\end{array}$

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### 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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