Integration Trigonometric Functions 

Integration of trigonometric functions involves finding the antiderivatives of functions like sine, cosine, tangent, and their reciprocals. This process is essential in calculus, as it helps solve problems related to areas, motion, and periodic phenomena. Common techniques include using standard formulas, substitution, and trigonometric identities to simplify the integrals. Mastery of integrating trigonometric functions is crucial for solving complex problems in physics, engineering, and other fields that involve waveforms, oscillations, and circular motion.

1.0What Are Trigonometric Functions?

Trigonometric functions, such as sine, cosine, tangent, and their reciprocals, are fundamental functions in mathematics that arise frequently in calculus and various applications. The process of integrating these functions involves finding their antiderivatives or primitive functions, which is crucial in solving many problems in physics, engineering, and other fields.

2.0Integration of Trigonometric Functions Using Substitution

One of the most powerful methods for integrating trigonometric functions is substitution. Substitution is particularly useful when the integrand involves composite trigonometric expressions or when there is a trigonometric identity that can simplify the expression.

Substitution Rule

When dealing with trigonometric integrals, substitution often involves recognizing a function and its derivative within the integrand. For example, if you have an integral of the form:

$\int f(g(x))g^{\prime }(x)dx$

You can substitute u = g(x), and consequently, du = g'(x)dx. This simplifies the integral and makes it easier to solve.

3.0Common Integration Trigonometric Functions Formulas

Here are some essential integration trigonometric functions formulas that you should be familiar with: 

$$\begin{gathered} 1.\text{Integral of Sine:} \\ \int \sin xdx=-\cos x+C \\ 2.\text{Integral of Cosine:} \\ \int \cos xdx=\sin x+C \\ 3.\text{Integral of Tangent:} \\ \int \tan xdx=-\ln |\cos x|+C \\ 4.\text{Integral of Secant Squared:} \\ \int {\sec }^{2}xdx=\tan x+C \\ 5.\text{Integral of Cosecant Squared:} \\ \int {\csc }^{2}xdx=-\cot x+C \\ 6.\text{Integral of Cotangent:} \\ \int \cot xdx=\ln |\sin x|+C \end{gathered}$$

These formulas are fundamental and are frequently used to solve various trigonometric integrals.

4.0Integration Trigonometric Functions Examples and Solutions

Example 1:

$(\int {\sin }^{2}xdx)$

Solution: 

To solve this, we can use the identity for ${\sin }^{2}x$:

Now, we integrate:

$$\begin{gathered} {\sin }^{2}x=\frac{1-\cos (2x)}{2} \\ \text{Now, we integrate:} \\ \int {\sin }^{2}xdx=\int \frac{1-\cos (2x)}{2}dx \\ =\frac{1}{2}\left[\int 1dx-\int \cos (2x)dx\right] \\ =\frac{x}{2}-\frac{\sin (2x)}{4}+C \end{gathered}$$

Example 2:

$\int {\sec }^{2}xdx$

Solution: 

We know from the formula that:

$\int {\sec }^{2}xdx=\tan x+C$

This is a straightforward integral, and the solution is $\tan x+C$.

Example 3:

$\int \sin x\cos xdx$

We can use the substitution method here. Notice that the product of sine and cosine suggests a substitution:

Let , $(u=\sin x),so(du=\cos xdx)$

Now, the integral becomes:

$\int udu=\frac{u^2}{2}+C=\frac{{\sin }^{2}x}{2}+C$

5.0Integration Trigonometric Functions in Class 12

For Class 12 students, integration of trigonometric functions is an essential topic in calculus. In this stage, students learn various techniques like substitution, integration by parts, and recognizing standard trigonometric identities. Mastering these concepts is vital not only for exams but also for real-world applications like physics and engineering.

Some key topics students should focus on include:

• Trigonometric identities: Recognizing and applying identities like $({\sin }^{2}x+{\cos }^{2}x=1),({\tan }^{2}x+1={\sec }^{2}x)$ , etc., helps simplify integrals.
• Substitution method: As shown in the examples above, substitution is a powerful tool for simplifying trigonometric integrals.
• Definite integrals: Students should also practice solving definite integrals involving trigonometric functions, which may require applying limits after integration.

6.0Solved Example of Integration Trigonometric Functions

Example 1:

$(\int \sin x\cos xdx)$

Solution:

$$\begin{gathered} \text{We can use the identity for the product of sine and cosine:} \\ \sin x\cos x=\frac{1}{2}\sin (2x) \\ \text{Now the integral becomes:} \\ \int \sin x\cos xdx=\frac{1}{2}\int \sin (2x)dx \\ \text{To integrate}(\sin (2x)),\text{we use the basic integral of sine:} \\ \int \sin (kx)dx=-\frac{1}{k}\cos (kx)+C \\ \text{So,} \\ \int \sin (2x)dx=-\frac{1}{2}\cos (2x)+C \\ \text{Thus, the solution is:} \\ \frac{1}{2}\left(-\frac{1}{2}\cos (2x)\right)+C=-\frac{1}{4}\cos (2x)+C \\ \text{So, the final result is:} \\ \int \sin x\cos xdx=-\frac{1}{4}\cos (2x)+C \end{gathered}$$

Example 2:

$(\int {\sin }^3(x)dx)$

Solution:

$$\begin{gathered} \text{We start by using the identity }{\sin }^2(x)=1-{\cos }^2(x) \\ \int {\sin }^3(x)dx=\int \sin (x)(1-{\cos }^2(x))dx \\ Now,let(u=\cos (x)),sothat(du=-\sin (x)dx).\text{Substituting into the integral:} \\ \int {\sin }^3(x)dx=-\int (1-u^2)du \\ \text{Now, integrate:}=-\int (1-u^2)du=-\left(u-\frac{u^3}{3}\right)+C \\ Substituteback(u=\cos (x)): \\ =-\left(\cos (x)-\frac{{\cos }^3(x)}{3}\right)+C \\ \text{Thus, the solution is:} \\ \int {\sin }^3(x)dx=-\cos (x)+\frac{{\cos }^3(x)}{3}+C \end{gathered}$$

Example 3:

$\int {\sec }^4(x)dx$

Solution:

$$\begin{gathered} \text{We can split}{\sec }^4(x))into({\sec }^2(x)\cdot {\sec }^2(x) \\ \int {\sec }^4(x)dx=\int {\sec }^2(x)\cdot {\sec }^2(x)dx \\ \text{Now, use the standard formula}(\int {\sec }^2(x)dx=\tan (x)),andlet(u=\tan (x)): \\ \int {\sec }^4(x)dx=\int {\sec }^2(x)(1+{\tan }^2(x))dx \\ Substitute(u=\tan (x))sothat(du={\sec }^2(x)dx),\text{and the integral becomes:} \\ \int (1+u^2)du=u+\frac{u^3}{3}+C \\ Substitutebacku=\tan (x): \\ \int {\sec }^4(x)dx=\tan (x)+\frac{{\tan }^3(x)}{3}+C \\ \text{Thus, the solution is:} \\ \int {\sec }^4(x)dx=\tan (x)+\frac{{\tan }^3(x)}{3}+C \end{gathered}$$

Example 4:

$\int {\tan }^2(x){\sec }^2(x)dx$

Solution:

$$\begin{gathered} \text{We know that}({\tan }^2(x)={\sec }^2(x)-1),so: \\ \int {\tan }^2(x){\sec }^2(x)dx=\int ({\sec }^2(x)-1){\sec }^2(x)dx \\ =\int {\sec }^4(x)dx-\int {\sec }^2(x)dx \\ \text{We already know that}(\int {\sec }^2(x)dx=\tan (x)),\text{so we only need to solve}(\int {\sec }^4(x)dx),\text{which we did earlier:} \\ \int {\sec }^4(x)dx=\tan (x)+\frac{{\tan }^3(x)}{3}+C \\ \text{Thus, the solution is:}\int {\tan }^2(x){\sec }^2(x)dx=\left(\tan (x)+\frac{{\tan }^3(x)}{3}\right)-\tan (x)+C \\ =\frac{{\tan }^3(x)}{3}+C \end{gathered}$$

Example 5:

$\int \frac{\sin (x)}{1+{\cos }^2(x)}dx$

Solution:

$$\begin{gathered} \text{Let’s use the substitution}u=\cos (x)sothatdu=-\sin (x)dx\text{The integral becomes:} \\ \int \frac{\sin (x)}{1+{\cos }^2(x)}dx=-\int \frac{du}{1+u^2} \\ \text{We recognize that} \\ \int \frac{1}{1+u^2}du={\tan }^{-1}(u), \\ so: \\ =-{\tan }^{-1}(u)+C \\ \text{Substitute back }u=\cos (x) \\ =-{\tan }^{-1}(\cos (x))+C \\ \text{Thus, the solution is:} \\ \int \frac{\sin (x)}{1+{\cos }^2(x)}dx=-{\tan }^{-1}(\cos (x))+C \end{gathered}$$

Example 6:

$\int \frac{{\sin }^2(x)}{{\cos }^3(x)}dx$

Solution:

$$\begin{gathered} \text{First, use the identity}{\sin }^2(x)=1-{\cos }^2(x) \\ \int \frac{{\sin }^2(x)}{{\cos }^3(x)}dx=\int \frac{1-{\cos }^2(x)}{{\cos }^3(x)}dx \\ \text{This simplifies to:} \\ =\int \frac{1}{{\cos }^3(x)}dx-\int \frac{1}{\cos (x)}dx \\ \text{Now, the first integral is }\int {\sec }^3(x)dx,\text{and the second is}\int \sec (x)dx.\text{Using known formulas:} \\ \int {\sec }^3(x)dx=\frac{1}{2}\sec (x)\tan (x)+\frac{1}{2}\ln |\sec (x)+\tan (x)|+C \\ And: \\ \int \sec (x)dx=\ln |\sec (x)+\tan (x)|+C \\ \text{Thus, the solution is:} \\ \int \frac{{\sin }^2(x)}{{\cos }^3(x)}dx=\frac{1}{2}\sec (x)\tan (x)+\frac{1}{2}\ln |\sec (x)+\tan (x)|-\ln |\sec (x)+\tan (x)|+C \end{gathered}$$

7.0Practice Question on Integration Trigonometric Functions

Question 1: Integrate $\int {\sin }^2(x)dx$

Try solving this integral using an identity and the appropriate integration formulas. (Hint: Use the identity _{\sin^2 x = \frac{1 - \cos(2x)}{2}} ).

Question 2: Integrate $\int {\tan }^2(x)dx$

Use the identity ${\tan }^2(x)={\sec }^2(x)-1$ to simplify and solve this integral.