# Basic integration

1. Jun 15, 2017

### Tanishq Nandan

1. The problem statement, all variables and given/known data

Integrate: 4cos(x/2).cos(x).sin(21x/2)
2. Relevant equations
▪sin(A+B)=sinAcosB+sinBcosA
▪2sinAcosA=sin2A
And obviously,
▪Integration of sinx is (-cosx)
▪Integration of cosx is (sinx)
3. The attempt at a solution
○I multiplied the numerator and denominator with sin(x/2)
○The numerator simplified to sin(2x)sin(21x/2) and in the denominator,we have sin(x/2).
○Now,I expanded sin(21x/2) by breaking 21x/2 into 10x and x/2 and using the above specified formula,which,on simplifying gives two terms.
○Now,the first term comes out to be sin2xcos10x,which can be integrated easily (well,maybe not one step but at least it can be done),but the second term is: sin(2x).sin(10x).cot(x/2)
How to integrate this term,it just eludes me..

2. Jun 16, 2017

### Fightfish

Well I wouldn't recommend this approach. You should instead consider the product-to-sum identities in order to simplify the product of trigonometric terms into a simple sum of individual sine and cosine terms. For example, $2 \cos \theta \cos \phi = \cos (\theta + \phi) + \cos (\theta - \phi)$.

3. Jun 16, 2017

### Tanishq Nandan

Oh..ok,got it.Thanks