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 sin(ax+b)sin(ax+b), cos(ax+b)cos(ax+b) and sec2(ax+b)sec^2(ax+b)

How to Integrate Trigonometric Functions

The simple integration forms of the trigonometric functions are as follows:

∫sinx dx=−cosx+C\int sinx\:dx = -cosx+C ∫cosx dx=sinx+C\int cosx\:dx = sinx+C ∫sec2x dx=tanx+C\int sec^2x\:dx=tanx+C\\

Where CC is some constant.

When written in the form of ax+bax+b the integration formulas are as follows:

∫sin(ax+b) dx=−1acos(ax+b)+C\int sin(ax+b)\:dx = -\frac{1}{a}cos(ax+b)+C ∫cos(ax+b) dx=1asin(ax+b)+C\int cos(ax+b)\:dx = \frac{1}{a}sin(ax+b)+C ∫sec2(ax+b) dx=1atan(ax+b)+C\\\int sec^2(ax+b)\:dx=\frac{1}{a}tan(ax+b)+C\\

Where CC is some constant.

Practice Question 1

Integrate the following function: y=cos(x2+3)y=cos(\frac{x}{2}+3)

Solution

Using the formula given above we can easily integrate the given equation as it is in the form cos(ax+b)cos(ax+b).

∫cos(x2+3) dx=∫cos(12x+3) dx=112sin(x2+3)+C=2sin(x2+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

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 xx is present in the integral. The general formula for the reverse chain rule can also be applied to the trigonometric functions.

∫undudx dx=un+1n+1+Cor∫(f(x))nf‘(x) dx=(f(x))n+1n+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

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 ∫02πsinx.cos4x.dx\int_{0}^{2\pi}sinx.cos^4x.dx

Solution

The first step in using the reverse-chain-rule is to define the variable uu and derive it.

u=cosxdudx=−sinxu = cosx\\\frac{du}{dx}=-sinx

Now, we must manipulate the integral to match that of the reverse-chain-rule formula.

∫dudx u4 dx=∫−sinxcos4x.dx=−∫sinxcos4x.dx\int \frac{du}{dx}\:u^4\:dx = \int -sinxcos^4x.dx\\=-\int sinxcos^4x.dx

Now that our answer is in the same form as the reverse-chain-rule formula, we can determine the answer.

−∫sinxcos4x dx=cos4+1x4+1+C=15cos5x+C-\int sinxcos^4x\:dx=\frac{cos^{4+1}x}{4+1}+C\\=\frac{1}{5}cos^5x+C

Previous

Differentiating Trigonometric Functions

Return to Module 6: The Trigonometric Functions