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
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
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)
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!
If you have any pending questions, please do go on to our Facebook page, Instagram or contact us directly at Math Lobby! We have certified mathematics tutors to aid you in your journey to becoming a better student!
As always: Work hard, stay motivated and we wish all students a successful and enjoyable journey with Math Lobby!
If you want to receive more Secondary Math Tips from us,
Visit our Website at: www.mathlobby.com
Visit and LIKE our Facebook page at www.facebook.com/mathlobbymotivation/
Visit and FOLLOW our Instagram page at https://www.instagram.com/mathlobbymotivation/
Join our Telegram group at https://t.me/mathlobby
Join our WhatsApp group at https://chat.whatsapp.com/IGhXU6lOoX83rDK7NMVvcw
Contact us via SMS/WhatsApp/Telegram/Call +65 96322202
*
*
*
#mathlobby#mathlobbymotivation#bestmathtuition#mathtuitionpotongpasir#mathtuitionsg#freemathresources#sgmaths#sgmathtutor#emath#amath#secschool#mathtutorsg #secondarymathtuition #nleveltuition#mathtuition#secondarymath#mathtutors#tuitionsg#singaporemath#sgtuition#sgstudents#sgmath