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# Integration- Trigonometry, Exponential and Logarithm

Updated: Jun 22, 2021 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 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