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iGCSE (2021 Edition)

12.05 Integration of trigonometric functions


Recall that we previously established the following results when differentiating trigonometric functions:

Differentiating trigonometric functions
$\frac{d}{dx}\sin x$ddxsinx $=$= $\cos x$cosx
$\frac{d}{dx}\cos x$ddxcosx $=$= $-\sin x$sinx


$\frac{d}{dx}\sin\left(ax+b\right)$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:

Integration of trigonometric functions
$\int\cos xdx$cosxdx $=$= $\sin x+C$sinx+C
$\int\sin xdx$sinxdx $=$= $-\cos x+C$cosx+C


$\int\cos\left(ax+b\right)dx$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 $a\ne0$a0


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.


$\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.


$\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



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$6sinxcosx.

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:

  1. Determine the value of $k$k.

  2. Now find an expression for $f\left(x\right)$f(x)

    You may use $C$C to represent an unknown constant.



Integrate functions of the form sin (ax + b) and cos (ax + b).

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