Calculus Tutorial
Trigonometric Integrals
Intro
Still in integration where our objective is to find the anti derivative of a certain function, in this case trigononetric functions. But first we will be dealing with simple trigonometric integrals.
Sample Problem
Solution
To do this simple integration, I am going to use u substitution.
Let u = 5x + 2
du = 5 dx
dx = du / 5
We are going to substitute it in the function,
S cos (5x + 2) dx = S cos u (du / 5)
Since there is a constant of 1/5 in du / 5 we will be transferring it outside the integral sign by using the formula
S a du = a S du where a is constant thus
S cos (5x + 2) dx = 1/5 S cos u du
Where S cos A dA = sin A + C
S cos (5x + 2) dx = 1/5 (sin u) + C
Since our answer is not final because we made u substitution thus replacing u with the original function
u = 5x + 2
S cos (5x + 2) dx = 1/5 (sin(5x + 2)) + C as your answer.
To verify if this is correct, check by differentiation.
Say y = 1/5 (sin (5x + 2)) + C
By using the formulas
d (c) = 0
d (cx) = c dx
d (sin u) = cos u du
y’ = 1/5 (cos (5x + 2)) (5)
We get the derivative of sine time the derivative of the angle. Cancelling 1/5 and 5 using multiplication thus,
y’ = cos (5x + 2)
Same as the given therefore making 1/5 (sin (5x + 2)) + C as the correct answer.
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