13.03 Antidiffferentiation and trigonometric functions
Recall that we previously established the following results when differentiating trigonometric functions:
| $$ddxsinx | $$= | $$cosx |
| $$ddxcosx | $$= | $$−sinx |
and
| $$ddxsin(ax+b) | $$= | $$acos(ax+b) |
| $$ddxcos(ax+b) | $$= | $$−asin(ax+b) |
Reversing these we get the following rules for integrating trigonometric functions:
| $$∫cosxdx | $$= | $$sinx+C |
| $$∫sinxdx | $$= | $$−cosx+C |
and
| $$∫cos(ax+b)dx | $$= | $$1asin(ax+b)+C |
| $$∫sin(ax+b)dx | $$= | $$−1acos(ax+b)+C |
Where $$a, $$b and $$C in each case are constants and $$a≠0
Worked examples
Example 1
Determine $$∫5cos(2x+π3)dx.
Think: To integrate we are going to divide by $$a from the term $$ax+b; here $$a=2. Then we change the function from cosine to sine.
Do:
$$∫5cos(2x+π3)dx=52sin(2x+π3)+C, where $$C is a constant.
Example 2
If $$f′(x)=0.5sin(x4) and $$f(2π)=−1, find $$f(x).
Think: We can first find the indefinite integral, and then use the given point $$(2π,−1) to find the value of the constant of integration.
For our integral, we will use the rule $$∫sin(ax+b)dx=−1acos(ax+b)+C. We have $$a=14, we want to divide by $$a, as well as change the sign and function. Remember, dividing by $$14 is the same as multiplying by $$4.
Do:
| $$∫0.5sin(x4)dx | $$= | $$−4×0.5cos(x4)+C |
Multiply by $$4, and change the function and sign |
| $$= | $$−2cos(x4)+C, where $$C is a constant |
Simplify |
Using the point $$(2π,−1), find $$C:
| $$f(2π) | $$= | $$−1 |
| $$∴−2cos(2π4)+C | $$= | $$−1 |
| $$0+C | $$= | $$−1 |
| $$C | $$= | $$−1 |
Thus, $$f(x)=−2cos(x4)−1.
Practice questions
Question 1
Integrate $$−5cos(x4).
You may use $$C as the constant of integration.
Question 2
State a primitive function of $$6sinx−cosx.
You may use $$C as a constant.
Question 3
Given that $$f′(x)=kcos3x, for some constant $$k, and that $$f′(0)=2 and $$f(π6)=6:
Determine the value of $$k.
Now find an expression for $$f(x)
You may use $$C to represent an unknown constant.