In this resource, you will find solved examples of triginometric examples. Before proceeding to examples and their solutions, first, let us see the anti derivatives of the common trigonometric functions.

Common Trigonometric Integrals

The integrals of trigonometric functions are referred to as trigonometric integrals. The integrals of the common trigonometric functions are compiled below:

1. \int cos x dx = sin x + C

2. \int sin x dx = -cos x + C

3. \int sec^2 x dx = tan x + C

4. \int cosec^2 x dx = - cotanx + C

5. \int (sec x tan x)dx = sec x + C

6. \int (cosec x cotan x)dx = - cosec x + C

 

The above list is quite helpful in solving the problems related to trigonometric integrals.

 

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Example 1

Calculate the following integral:

\int (4 - sin x) dx

First, use the sum/difference property of integration to write the above function like this:

= \int 4 dx - \int sin x dx

Compute the integrals of both the terms separately. The integral of 4 is 4x and the integral of sin x is cos x:

= 4x - cos x + C

Example 2

Calculate the following integral:

\int sin (4x + 5) dx

Solution

Using the sum/difference property of integration, we can write the above function as:

= \int 4x dx + \int 5 dx

Calculate the integral or antiderivative of the two terms separately. The antiderivative of 4x is equal to 2x^2 and the antiderivative of 5 is equal to 5x:

= 2x^2 + 5x + C

E

Example 3

Compute the antiderivative of the following function:

\int (3 cosec^2 x + 5 cos x) dx

Solution

\int (3 csc^2 x + 5 cos x) dx can be written as:

= \int 3 csc^2 x + \int 5 cos x dx

According to one of the properties of integrals, we can shift constants before the integral sign. Hence, we can write the above function like this:

= 3\int  csc^2 x + 5 \int  cos x dx

We know that \int csc^2 x dx = - cotan x + C and \int cos x dx = sin x + C. Hence, we will substitute these values in the above function like this:

= - 3 cotan x + 5 sin x + C

Example 4

Calculate the integral of the following function:

\int sin (9x + 7) dx

Solution

In this problem, we will use substitution to compute the integral. Suppose 9x + 7 = u.

If u = 9x + 7, then \frac{du}{dx} = 9. This means that dx  is equal to \frac{1}{9} du. Now, we will substitute these values in the original function to get the following function:

= \int sin u \frac{1}{9} du

Shift the fraction before the integral sign:

= \frac{1}{9} \int sin u du

We know that \int sin x dx = -cos x + C. Hence, we will substitute this value in the above function:

= -\frac{1}{9} cos u + C

Since u = 9x + 7, hence we will substitute this value of u in the above function again to get the final answer:

= \frac{1}{9} cos (9x + 7) + C

 

Example 5

Calculate the integral of the following function:

\int cos (6x + 1) dx

Solution

In this problem, we will use substitution to compute the integral. Suppose 6x + 1 = u

If u = 6x + 1, then \frac{du}{dx} = 6. This means that dx  is equal to \frac{1}{6} du. Now, we will substitute these values in the original function to get the following function:

= \int cos u \frac{1}{6} du

Shift the fraction before the integral sign:

= \frac{1}{6} \int cos u du

We know that \int cos x dx = sin x + C. Hence, we will substitute this value in the above function:

= \frac{1}{6} sin u + C

Since u = 6x + 1, hence we will substitute this value of u in the above function again to get the final answer:

= \frac{1}{6} sin (6x + 1) + C

Example 6

Calculate the following integral:

\int x^3 sinx^4dx

Solution

Suppose u = x^4, then du = 4x^3dx. It means that x^3dx is equal to \frac{du}{4}.

= \int sin u \frac{du}{4}

Shift fraction on the left side of the integral to get:

= \frac{1}{4} \int sin u du

Remember that the antiderivative of sin u is equal to - cos u + C. Hence, we can write the function as:

= \frac{1}{4} (-cos u) + C

Put u = x^4 to get the following answer:

= -\frac{1}{4} cosx^4 + C

 

Example 7

Calculate the following integral:

\int x^2 cos x^3dx

Solution

Suppose u = x^3, then du = 3x^2dx. It means that x^3dx is equal to \frac{du}{3}.

= \int sin u \frac{du}{3}

Shift fraction on the left side of the integral to get:

= \frac{1}{3} \int cos  u du

Remember that the antiderivative of cos u is equal to sin u + C. Hence, we can write the function as:

= \frac{1}{3} (sin u) + C

Put u = x^3 to get the following answer:

= \frac{1}{3} (sinx^3) + C

 

Example 8

Calculate the integral of the following function:

\int sin^3 x dx

Solution

We can write sin^3x as the product of sin^2 x and sin x like this:

= \int sin^2x \cdot sin x

We know that sin^x + cos^2 x = 1. This means that sin^2 x is equal to 1 - cos^2 x.

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

Substitute u = cos x:

= \int (1 - u^2) \cdot sin u du

\int 1 - u^2 is equal to u - \frac{u^3}{3} :

= u - \frac{u^3}{3}

Substitute u = cos x:

= cos x - \frac{cos^3 x}{3} + C

 

Example 9

Calculate the integral of the following function:

\int tan (3x - 4) dx

Solution

In this problem, we will use substitution to compute the integral. Suppose 3x - 4 = u

If u = 3x - 4, then \frac{du}{dx} = 3. This means that dx  is equal to \frac{1}{3} du. Now, we will substitute these values in the original function to get the following function:

= \int tan u \frac{1}{3} du

Shift the fraction before the integral sign:

= \frac{1}{3} \int tan u du

We know that \int tan x dx = - ln |cos x| + C. Hence, we will substitute this value in the above function:

= \frac{1}{3} -ln |cos u| + C

Since u = 3x - 4, hence we will substitute this value of u in the above function again to get the final answer:

= -\frac{1}{3}  ln |cos 3x - 4| + C

 

Example 10

Evaluate the following function:

\int cos ^ 4 x  sin^3 x dx

Solution

As the power of sin x is odd, hence we can write the above function like this:

= \int cos ^4 x sin^2x sin x dx

Substitute sin^2 x = 1 - cos^2 x in the above function:

= \int cos^4x (1 - cos^2x) sin x dx

Suppose u = cos x, then du = -sin x:

= \int u^4 (1 - u^2x) (-du)

= \int - (u^4 - u^8)

= \int - u^4 + u^8

Integrate the above function like this:

= - \frac{1}{5} u^5 + \frac{1}{9} u^9 + C

Substitute u = cos x again in the above function:

= - \frac{1}{5} cos ^5 x + \frac{1}{9} cos^9 x + C

 

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Rafia Shabbir