__In this note, you will learn:__

· __Integration of trigonometric functions__

· __Integration of exponential functions__

· __Integration of functions in the form of 1/x and 1/ (ax + b)__

__Integration of trigonometric functions__

__Integration of trigonometric functions__

Under the chapter of ** differentiation of trigonometric functions**, we can recall that:

**d/dx [sin(x)] = cos(x)**

**d/dx [cos(x)] = -sin(x)**

**d/dx [tan(x)] = sec^2(x)**

Since we are aware that integration is the reverse operation of differentiation, we can hence deduce that:

**∫ cos x dx = sin x + c,**

**∫ sin x dx = -cos x + c,**

**∫ sec^2 x dx = tan + c,**

**where x is in degrees and radians while c is an arbitrary constant**

For both differentiation and integration of trigonometric functions, it is only done when the angles involved in the argument of the function are measured in radians.

If **a** and **b** are constants, since **d/dx [sin (ax + b)] = a cos (ax + b) + c**,

**∫ a cos (ax + b) dx = sin (ax + b) + c,**

**∫ cos (ax + b) dx = (1/a) sin (ax + b) + c,**

**where c is an arbitrary constant.**

From the explanation above, we can attain similar results if we were to integrate trigonometric functions involving sine and secant squared. Therefore, we will get:

**∫ cos (ax + b) dx = (1/a) sin (ax + b) + c,**

**∫ sin (ax + b) dx = (-1/a) cos (ax + b) + c,**

**∫ sec^2 (ax + b) dx = (1/a) tan (ax + b) + c,**

**where x is in degrees and radians, a and b are constants, and c is an arbitrary constant**

Let’s take a look at how we can apply these integration formulae in some examples below:

Integrate each of the following with respect to x.

a) 3 sin 4x + (1/2) cos 3x

b) 5/ (3 cos^2 6x)

c) tan^2 x

Solution:

a) ∫ [3 sin 4x + (1/2) cos 3x] dx

Note: Recall the two rules that we have learnt back from the Introduction to Integration, we are able to split the sine and cosine terms using the **Addition Rule of Integration**, and followed by the **Scalar Multiple Rule** for the terms individually.

= ∫ (3 sin 4x) dx + ∫ [(1/2) cos 3x] dx

= 3 ∫ (sin 4x) dx + (1/2) ∫ (cos 3x) dx

= 3 [(-1/4) (cos 4x)] + (1/2) [(1/3) (sin 3x)] + c

= (-3/4) cos 4x + (1/6) sin 3x + c

b) ∫ [5/ (3 cos^2 6x)] dx

= (5/3) ∫ (1/ cos^2 6x) dx **(Scalar Multiple Rule)**

= (5/3) ∫ (sec^2 6x) dx

**Note: The reciprocal of a cosine function = secant function, therefore 1/ cos (ax + b) = sec (ax + b)**.

= (5/3) [(1/6) (tan 6x)] + c

= (5/18) tan 6x + c

c) ∫ tan^2 x dx = ∫ (sec^2 x – 1) dx

= tan x – x + c

Note: DO NOT solve **∫ tan^2** **x** dx like this: **∫ tan^2 x dx = (1/3) (tan^3 x) + c**

More questions involving trigonometric functions with argument in the form of (ax + b):

Find each of the following integrals.

a) ∫ [x^3 + sec^2 (8 – 3x)] dx

b) ∫ sin [(7/8) x + 5] dx

c) ∫ (2/5) cos (π – 3x) dx

Solution:

a) ∫ [x^3 + sec^2 (8 – 3x)] dx

= ∫ x^3 dx + ∫ sec^2 (8 – 3x) dx

= [(1/4) x^4] + [(-1/3) tan (8 – 3x)] + c

= (1/4) x^4 – (1/3) tan (8 – 3x) + c

b) ∫ sin [(7/8) x + 5] dx

= [-1/ (7/8)] cos [(7/8) x + 5] + c

= (-8/7) cos [(7/8) x + 5] + c

c) ∫ (2/5) cos (π – 3x) dx

= (2/5) ∫ cos (π – 3x) dx

= (2/5) [(-1/3) sin (π – 3x)] + c

= (-2/15) sin (π – 3x) + c

__Integration of exponential functions__

__Integration of exponential functions__

Under the chapter of ** differentiation of exponential functions**, we can recall that:

**d/dx (e^x) = e^x**

**d/dx (e^(ax + b)) = a e^(ax + b)**

Since we are aware that integration is the reverse operation of differentiation, we can hence deduce that:

**∫ ex dx = e^x + c,**

**∫ e^(ax + b) dx = (1/a) e^(ax + b) + c,**

**where a and b are constants, and c is an arbitrary constant**

When we are dealing with integrals, we must have the full clarity of whether the base or the index is the variable. We can see from the two examples shown below:

**∫ x^n dx = [(x^(n+1))/ (n+1)] + c**

As you can see, the **variable**, **x**, in this integral is at the **base** with the **index**, **n**, as the constant. Hence, this will be an integration with regards to a power function, therefore we must use the respective formula for the integration of a **power function**.

**∫ e^x dx = e^x + c**

Whereas in this case, where the base is a constant, **e**, and the **variable**, **x**, is the **index**, we must hence use the formula for the integration of an **exponential function**, which is **vastly different** from the formula for the integration of a power function.

Let’s take a look at how we can apply these integration formulae for exponential functions in some examples below:

Find each of the following integrals.

a) ∫ 4e^5x dx

b) ∫ e^(3 – 4x) dx

Solution:

a) ∫ 4e^5x dx

= 4 ∫ e^5x dx

= 4 [(1/5) e^5x] + c

= (4/5) e^5x + c

b) ∫ e^(3 – 4x) dx

= (-1/4) e^(3 – 4x) dx

__Integration of functions in the form of 1/x and 1/ (ax + b)__

__Integration of functions in the form of 1/x and 1/ (ax + b)__

Referring back to the article, “** Introduction to Integration**”, we know that:

**∫ x^n dx = [(x^(n+1))/ (n+1)] + c,**

**where n ≠ -1**

It is because if **n** were to be -1, the formula above will not be valid since the denominator, **n + 1**, will be 0 and we cannot divide anything by zero as it will be deemed as **undefined**.

Hence, how do we solve for the integral of **x-1**, or **1/x**?

Under the chapter of ** differentiation of logarithmic functions**, we can recall that:

**d/dx (ln x) = 1/x,**

**where x > 0, and**

**d/dx [ln (ax + b)] = a [1/ (ax + b)] = a/ (ax + b),**

**where (ax + b) > 0**

Since we are aware that integration is the reverse operation of differentiation, we can hence deduce that:

**∫ (1/x) dx = ln x + c,**

**where x > 0,**

**∫ [1/ (ax + b)] dx = (1/a) ln (ax + b) + c,**

**where ax + b > 0, a and b are constants and c is an arbitrary constant**

Let’s take a look at how we can apply these integration formulae in some examples below:

Integrate each of the following with respect to x.

a) 3/4x

b) 1/ (3x + 5)

c) 9/ (6 – 2x)

Solution:

a) ∫ (3/4x) dx

= 3 ∫ (1/4x) dx

= 3 (1/4) ln 4x + c

= (3/4) ln 4x + c

b) ∫ [1/ (3x + 5)] dx

= (1/3) ln (3x + 5) + c

c) ∫ [9/ (6 – 2x)] dx

= 9 ∫ [1/ (6 – 2x)] dx

= 9 (1/-2) ln (6 – 2x) + c

= (-9/2) ln (6 – 2x) + c

And that’s all for today, students! ** Math Lobby** hopes that after this article, you have a clear understanding on the integration of trigonometric, exponential and logarithmic functions and is equipped with the necessary skills to deal with questions involving the application of the formulae you have learnt above!

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