Integrating Trig Functions (Cambridge International IGCSE Further Maths)

Integrating trig functions

How do I integrate sin, cos and sec^2?

• The integrals for sine and cosine are

$\int \sin x d x = - \cos x + c$

$\int \cos x d x = \sin x + c$

where$c$ is the constant of integration

• Also, from the derivative of$\mathrm{tan}x$

$\text{integral} \text{sec} ^{ } 2 x d x = \text{tan} x + c$

• For the linear function$ax+b$, where$a$ and$b$ are constants,

$\text{integral} \text{sin} ( a x + b ) d x = - \frac{1}{ } a \text{cos} ( a x + b ) + c$

$\text{integral} \text{cos} ( a x + b ) d x = \frac{1}{ } a \text{sin} ( a x + b ) + c$

$\text{integral} \text{sec} ^{ } 2 ( a x + b ) d x = \frac{1}{ } a \text{tan} ( a x + b ) + c$

• For calculus with trigonometric functions angles must be measured in radians

Exam Hint

• Remember to add 'c', the constant of integration, for any indefinite integrals

Worked Example a) Find, in the form$f(x)+c$, an expression for each integral

i. $\int \cos x \mathrm{d} x$

ii. integral sec ^2 stretchy ( 3 x - pi / 3 stretchy ) d x

i. This is a result you should be able to recall from memory.

$\int \cos x \mathrm{d} x = \sin x + c$

ii. Use the standard result: $\int \sec ^{2 } ( a x + b ) d x = \frac{ 1}{a} \tan ( a x + b ) + c .$

equation

b) A curve has the following equation:

y = integral 2 sin ( 2 x + pi / 6 ) d x

The curve passes through the point with coordinates ( pi / 3 , sqrt( 3 ).

Find an expression for $y.$

Factor out the 2 in front of the sin first, and then use the standard result: $\int \sin ( a x + b ) d x = - \frac{1}{a} \cos ( a x + b ) + c .$* *

y = 2 integral sin ( 2 x + pi / 6 ) d x > y = 2 ( - 1 half cos ( 2 x + pi / 6 ) + c )

When multiplying the arbitrary constant $c$, by 2, it is still just an arbitrary constant and so it is still just written as $c.$* *

y = - cos ( 2 x + pi / 6 ) + c

Substitute in the given coordinate, and evaluate, to find $c.$* *

At $x=\frac{π}{3},y=\sqrt{3}$

sqrt(3=- cos ((2π)/(3)+π / 6)+c>sqrt(3=- cos ((5π)/(6))+c>sqrt(3=(sqrt(3)/(2)+c>c=(sqrt(3)/(2)

Rewrite the full expression for $y.$

$y=-\mathrm{cos}(2x+\frac{π}{6})+\frac{\sqrt{3}}{2}$