Integration Using Double Angle Formulae
In order to integrate 
, for example, it might be tempting to use the basic trigonometric identity 
as this identity is more familiar. However, integrating 
is more complicated than integrating itself. Instead, we can use a double angle identity to integrate . Recall the double angle formulae:
and 
.
The final identity for 
can be rearranged to get an expression for . That is, 
. This is easy to integrate and so:
A similar process can be applied to integrate 
. We can also integrate when the argument is different – see Example 1. To integrate 
, we use a reciprocal trigonometric identity. See below.
Using Reciprocal Trigonometric Identities
We can use the double angle formulae to integrate and . However, for integrating we may use a reciprocal trigonometric identity. Recall that 
. Hence,
using the formula booklet to integrate 
. See Example 2.
Using Compound Angle Formulae
It is important to remember, as well as the above, that a question may ask you to integrate a trigonometric function which, at first, looks hugely unfamiliar. For example, how could we integrate 
? It might be tempting to try integration by parts since it is a product. However, the formula booklet provides compound angle identities that will prove useful in integrating this kind of function:
.
The first formula will help us to integrate . We rewrite the formula as 
. The individual terms on the right of this will be much easier to integrate than the product on the left. Hence, we must find 
and 
such that 
and 
. It follows that 
and 
. Hence, 
and 
. The integral becomes:
Note that or could be negative and we would have to use the odd/even properties of sin/cos – see Example 3. It is possible to integrate using integration by parts but it is much simpler to use the method above. See Integration by Parts Example 3.
Examples
Integrate 
.
Solution:
Recall the double angle formula 
. This can be rewritten as 
. It follows that 
. Hence,
Find the exact value of 
.
Solution:
Firstly, expanding the brackets gives:
where we transformed terms into those that we can integrate. See Example 1 for more on transforming . Hence,
![\begin{array}{lll}\int_0^\pi\left(\tan(x)+\cos(x)\right)^2&=&\int_0^\pi\left(\tan^2(x)+2\sin(x)+\frac{1}{2}\cos(2x)+\frac{1}{2}\right)dx\\&=&\left[\tan(x)-x-2\cos(x)+\frac{1}{4}\sin(2x)+\frac{1}{2}x\right]_0^\pi\\&=&\left[\tan(x)-2\cos(x)+\frac{1}{4}\sin(2x)-\frac{1}{2}x\right]_0^\pi\\&=&\left(0+2+0-\frac{1}{2}\pi\right)-\left(0-2+0-0\right)\\&=&4-\frac{1}{2}\pi\end{array}](https://studywell.com/wp-content/ql-cache/quicklatex.com-7e9da117692409bf6e6b0098996245c0_l3.png "Rendered by QuickLaTeX.com")
Find 
.
Solution:
Using the formula 
we can write 
. It follows that we must find and such that 
and 
or 
and 
. So, and 
. Hence, 
. Note that cos is an even function (see more on odd/even functions) and so 
. The integral becomes:
