Integration of Trigonometric Functions
Expert reviewed • 08 January 2025 • 5 minute read
HSC Maths Advanced Syllabus
establish and use the formulae for the anti-derivatives of s i n ( a x + b ) sin(ax+b) s in ( a x + b ) c o s ( a x + b ) cos(ax+b) cos ( a x + b ) s e c 2 ( a x + b ) sec^2(ax+b) se c 2 ( a x + b )
VIDEO
How to Integrate Trigonometric Functions
The simple integration forms of the trigonometric functions are as follows:
∫ s i n x d x = − c o s x + C \int sinx\:dx = -cosx+C ∫ s in x d x = − cos x + C ∫ c o s x d x = s i n x + C \int cosx\:dx = sinx+C ∫ cos x d x = s in x + C ∫ s e c 2 x d x = t a n x + C \int sec^2x\:dx=tanx+C\\ ∫ se c 2 x d x = t an x + C Where C C C
When written in the form of a x + b ax+b a x + b
∫ s i n ( a x + b ) d x = − 1 a c o s ( a x + b ) + C \int sin(ax+b)\:dx = -\frac{1}{a}cos(ax+b)+C ∫ s in ( a x + b ) d x = − a 1 cos ( a x + b ) + C ∫ c o s ( a x + b ) d x = 1 a s i n ( a x + b ) + C \int cos(ax+b)\:dx = \frac{1}{a}sin(ax+b)+C ∫ cos ( a x + b ) d x = a 1 s in ( a x + b ) + C ∫ s e c 2 ( a x + b ) d x = 1 a t a n ( a x + b ) + C \\\int sec^2(ax+b)\:dx=\frac{1}{a}tan(ax+b)+C\\ ∫ se c 2 ( a x + b ) d x = a 1 t an ( a x + b ) + C Where C C C
Practice Question 1
Integrate the following function: y = c o s ( x 2 + 3 ) y=cos(\frac{x}{2}+3) y = cos ( 2 x + 3 )
Using the formula given above we can easily integrate the given equation as it is in the form c o s ( a x + b ) cos(ax+b) cos ( a x + b )
∫ c o s ( x 2 + 3 ) d x = ∫ c o s ( 1 2 x + 3 ) d x = 1 1 2 s i n ( x 2 + 3 ) + C = 2 s i n ( x 2 + 3 ) + C \int cos(\frac{x}{2}+3)\:dx = \int cos(\frac{1}{2}x+3)\:dx\\=\frac{1}{\frac{1}{2}}sin(\frac{x}{2}+3)+C\\=2sin(\frac{x}{2}+3)+C ∫ cos ( 2 x + 3 ) d x = ∫ cos ( 2 1 x + 3 ) d x = 2 1 1 s in ( 2 x + 3 ) + C = 2 s in ( 2 x + 3 ) + C View Solution Using the Reverse Chain Rule
As we have discussed in previous modules, the reverse chain rule is a tool used in integration, to assist when more than one value of x x x
∫ u n d u d x d x = u n + 1 n + 1 + C o r ∫ ( f ( x ) ) n f ‘ ( x ) d x = ( f ( x ) ) n + 1 n + 1 + C \int u^n\frac{du}{dx}\:dx = \frac{u^{n+1}}{n+1}+C \qquad or \qquad \int (f(x))^n f^`(x)\:dx=\frac{(f(x))^{n+1}}{n+1}+C ∫ u n d x d u d x = n + 1 u n + 1 + C or ∫ ( f ( x ) ) n f ‘ ( x ) d x = n + 1 ( f ( x ) ) n + 1 + C When we use this formula, the simplest approach to solving a question is to first find and define all variables, then substitute their values.
Practice Question 2
Solve ∫ 0 2 π s i n x . c o s 4 x . d x \int_{0}^{2\pi}sinx.cos^4x.dx ∫ 0 2 π s in x . co s 4 x . d x
The first step in using the reverse-chain-rule is to define the variable u u u
u = c o s x d u d x = − s i n x u = cosx\\\frac{du}{dx}=-sinx u = cos x d x d u = − s in x Now, we must manipulate the integral to match that of the reverse-chain-rule formula.
∫ d u d x u 4 d x = ∫ − s i n x c o s 4 x . d x = − ∫ s i n x c o s 4 x . d x \int \frac{du}{dx}\:u^4\:dx = \int -sinxcos^4x.dx\\=-\int sinxcos^4x.dx ∫ d x d u u 4 d x = ∫ − s in x co s 4 x . d x = − ∫ s in x co s 4 x . d x Now that our answer is in the same form as the reverse-chain-rule formula, we can determine the answer.
− ∫ s i n x c o s 4 x d x = c o s 4 + 1 x 4 + 1 + C = 1 5 c o s 5 x + C -\int sinxcos^4x\:dx=\frac{cos^{4+1}x}{4+1}+C\\=\frac{1}{5}cos^5x+C − ∫ s in x co s 4 x d x = 4 + 1 co s 4 + 1 x + C = 5 1 co s 5 x + C View Solution