Integrating trigonometric functions involving brackets and powers
We have seen from integrating simple trigonometric expressions that in general:
\(\int {\sin x\,dx} \to - \cos x + c\)
and:
\(\int {\cos x\,dx} \to \sin x + c\)
Note: This only works when \(x\) is measured in radians.
We are now going to look at more complex trigonometric functions where we will use the general rule:
\(\int {\cos (ax + b)dx = \frac{1}{a}} \sin (ax + b) + c\)
\(\int {\sin (ax + b)dx = - \frac{1}{a}} \cos (ax + b) + c\)
Again, this only works when \(x\) is measured in radians.
Question
Extension
Find \(\int {\cos 6x\,\,dx}\)
\(\int {\cos 6x\,\,dx}\)
\(= \frac{1}{6}\sin 6x + c\)
Question
Extension
Find \(\int {\sin (2 - 3x)\,\,dx}\)
\(\int {\sin (2 - 3x)\,\,dx}\)
\(= \frac{{ - 1}}{{ - 3}}\cos (2 - 3x) + c\)
\(= \frac{1}{3}\cos (2 - 3x) + c\)
Question
Extension
Find \(\int {2\cos (1 - 6x)\,\,dx}\)
\(\int {2\cos (1 - 6x)\,\,dx}\)
\(= \frac{2}{{ - 6}}\sin (1 - 6x) + c\)
\(= - \frac{1}{3}\sin (1 - 6x) + c\)
We can also use the rules we have learned to integrate a trigonometric expression with powers.
In order to do this, we need to express it in a different form first before we integrate.
For example, if we were asked to integrate:
\(\int {{{\sin }^2}} x\,\,dx\)
Using our knowledge from the double angle formulae, we know that \({\sin ^2}x = \frac{1}{2}(1 - \cos 2x)\), so this can be found by:
\(\int\frac{1}{2}(1-cos2x)\,\,dx\)
\(=\int (\frac{1}{2}-\frac{1}{2}cos2x)\,\,dx\)
\(= \frac{1}{2}x - \frac{1}{2} \times \frac{1}{2}\sin 2x + c\)
\(= \frac{1}{2}x - \frac{1}{4}\sin 2x + c\)
