# Homework Help: Integration Question

1. Oct 30, 2005

### Selectron

I'm stuck half-way through a question where I need to find the integral (with respect to x) of:

tan(x/2).(cosec x + cot x)^2

I've tried every method of integration that I know and I can't get it. Is it actually possible to integrate the expression?

If it's any help, this problem is part of the further maths A-level and the original problem was:

Find the general solution of the differential equation dy/dx - 2y.cosec x = tan(x/2), 0<x<pi

I got to the integral I'm stuck on by using an integrating factor of (cosec x + cot x)^2, so that the LHS became the derivative of y(cosec x + cot x)^2 (and therefore y(cosec x + cot x)^2 = integral[tan(x/2).(cosec x + cot x)^2]dx ).

P.S. Sorry if the above isn't very clear, since I don't know how to get mathematical symbols on these boards.

2. Oct 30, 2005

### benorin

The value of that integral is 2ln(sin(x/2))+C

All you need to know is the right trig identity: $$\tan \left(\frac{x}{2}\right)=\frac{\sin(x)}{\cos(x)+1}$$.

3. Oct 30, 2005

### Fermat

The integating factor simplifies by using half-angle formulae.

sinx = 2sin(x/2).cos(x/2)
cos(x) = 2cos²(x/2) - 1

btw, you can get maths symbols by using Latex

4. Oct 30, 2005

### benorin

Here's the integral

Use $$\tan\left(\frac{x}{2}\right)=\frac{\sin(x)}{\cos(x)+1}$$ like this:
$$\int\tan\left(\frac{x}{2}\right)\left(\csc(x) + \cot(x)\right)^{2}dx = \int\left(\frac{\sin(x)}{\cos(x)+1}\right)\left(\frac{1}{\sin(x)} + \frac{\cos(x)}{\sin(x)}\right)^{2}dx$$
$$= \int\left(\frac{\sin(x)}{\cos(x)+1}\right)\left(\frac{\cos(x)+1}{\sin(x)}\right)^{2}dx = \int\frac{\cos(x)+1}{\sin(x)}dx = \int\cot\left(\frac{x}{2}\right)dx$$
Let $$u=\frac{x}{2}\Rightarrow 2du=dx$$ to get
$$2\int\cot(u)du = 2\int\frac{\cos(u)}{\sin(u)}du$$
set $$w=\sin(u)\Rightarrow dw=\cos(u)du$$ which yeilds
$$2\int\frac{dw}{w} = 2\ln(w) + C = 2\ln(\sin(u)) + C = 2\ln\left(\sin\left(\frac{x}{2}\right)\right) + C$$
There, all done.

Last edited: Oct 30, 2005
5. Nov 3, 2005

### Selectron

Thanks guys. You've been very helpful.