13.03 Antidiffferentiation and trigonometric functions
Recall that we previously established the following results when differentiating trigonometric functions:
| $\frac{d}{dx}\sin x$ddxsinx | $=$= | $\cos x$cosx |
| $\frac{d}{dx}\cos x$ddxcosx | $=$= | $-\sin x$−sinx |
and
| $\frac{d}{dx}\sin\left(ax+b\right)$ddxsin(ax+b) | $=$= | $a\cos\left(ax+b\right)$acos(ax+b) |
| $\frac{d}{dx}\cos\left(ax+b\right)$ddxcos(ax+b) | $=$= | $-a\sin\left(ax+b\right)$−asin(ax+b) |
Reversing these we get the following rules for integrating trigonometric functions:
| $\int\cos xdx$∫cosxdx | $=$= | $\sin x+C$sinx+C |
| $\int\sin xdx$∫sinxdx | $=$= | $-\cos x+C$−cosx+C |
and
| $\int\cos\left(ax+b\right)dx$∫cos(ax+b)dx | $=$= | $\frac{1}{a}\sin\left(ax+b\right)+C$1asin(ax+b)+C |
| $\int\sin\left(ax+b\right)dx$∫sin(ax+b)dx | $=$= | $-\frac{1}{a}\cos\left(ax+b\right)+C$−1acos(ax+b)+C |
Where $a$a, $b$b and $C$C in each case are constants and $a\ne0$a≠0
Worked examples
Example 1
Determine $\int5\cos\left(2x+\frac{\pi}{3}\right)dx$∫5cos(2x+π3)dx.
Think: To integrate we are going to divide by $a$a from the term $ax+b$ax+b; here $a=2$a=2. Then we change the function from cosine to sine.
Do:
$\int5\cos\left(2x+\frac{\pi}{3}\right)dx=\frac{5}{2}\sin\left(2x+\frac{\pi}{3}\right)+C$∫5cos(2x+π3)dx=52sin(2x+π3)+C, where $C$C is a constant.
Example 2
If $f'\left(x\right)=0.5\sin\left(\frac{x}{4}\right)$f′(x)=0.5sin(x4) and $f\left(2\pi\right)=-1$f(2π)=−1, find $f\left(x\right)$f(x).
Think: We can first find the indefinite integral, and then use the given point $\left(2\pi,-1\right)$(2π,−1) to find the value of the constant of integration.
For our integral, we will use the rule $\int\sin\left(ax+b\right)dx=-\frac{1}{a}\cos\left(ax+b\right)+C$∫sin(ax+b)dx=−1acos(ax+b)+C. We have $a=\frac{1}{4}$a=14, we want to divide by $a$a, as well as change the sign and function. Remember, dividing by $\frac{1}{4}$14 is the same as multiplying by $4.$4.
Do:
| $\int0.5\sin\left(\frac{x}{4}\right)dx$∫0.5sin(x4)dx | $=$= | $-4\times0.5\cos\left(\frac{x}{4}\right)+C$−4×0.5cos(x4)+C |
Multiply by $4$4, and change the function and sign |
| $=$= | $-2\cos\left(\frac{x}{4}\right)+C$−2cos(x4)+C, where $C$C is a constant |
Simplify |
Using the point $\left(2\pi,-1\right)$(2π,−1), find $C$C:
| $f\left(2\pi\right)$f(2π) | $=$= | $-1$−1 |
| $\therefore-2\cos\left(\frac{2\pi}{4}\right)+C$∴−2cos(2π4)+C | $=$= | $-1$−1 |
| $0+C$0+C | $=$= | $-1$−1 |
| $C$C | $=$= | $-1$−1 |
Thus, $f\left(x\right)=-2\cos\left(\frac{x}{4}\right)-1$f(x)=−2cos(x4)−1.
Practice questions
Question 1
Integrate $-5\cos\left(\frac{x}{4}\right)$−5cos(x4).
You may use $C$C as the constant of integration.
Question 2
State a primitive function of $6\sin x-\cos x$6sinx−cosx.
You may use $C$C as a constant.
Question 3
Given that $f'\left(x\right)=k\cos3x$f′(x)=kcos3x, for some constant $k$k, and that $f'\left(0\right)=2$f′(0)=2 and $f\left(\frac{\pi}{6}\right)=6$f(π6)=6:
Determine the value of $k$k.
Now find an expression for $f\left(x\right)$f(x)
You may use $C$C to represent an unknown constant.