## 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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