Hope this helps!Best regards,Nalin Pithwa

[2^Oscar]

Hi,

I've been doing some additional maths papers and I've seen the use of the substitution u=tan(x/2) in order to calculate integrals. In the mark scheme it states that this particular substitution used to be fairly common, however is not on the modern A-level syllabus.

Would someone please mind advising me of suitable situations to use such a substitution? I am struggling to see when I should use it.


Thanks,
Oscar

 

[arildno]

It is useful when your integrand consists of a ratio between to polynomials in of trignometric functions.

For example:

Let's look at:
$$\int\frac{\cos\theta}{2\cos\theta-\sin\theta}d\theta$$

How would you integrate that one?

Not very easy, but look at the following:
$$\cos\theta=\cos^{2}\frac{\theta}{2}-\sin^{2}\frac{\theta}{2}=\cos^{2}\frac{\theta}{2}(1-\tan^{2}\frac{\theta}{2})=\frac{1-u^{2}}{1+u^{2}}$$
$$\sin\theta=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}=2\cos^{2}\frac{\theta}{2}\tan\frac{\theta}{2}=\frac{2u}{1+u^{2}}$$

Thus, we also have:
$$\tan\theta=\frac{2u}{1-u^{2}}$$

In addition, we have:
$$\frac{du}{d\theta}=\frac{1}{2}\frac{1}{\cos^{2}\frac{\theta}{2}}=\frac{1}{2}(u^{2}+1}\to{d\theta}=\frac{2du}{1+u^{2}}$$

Thus, the above integral can be converted into a rational expression of polynomials in the variable "u", and that can be solved using partial fractions decomposition.