# Factorizing denominator help

1. Jun 21, 2005

### Yegor

$$\int\frac{1}{(\sin(x)+\frac{2}{\cos(x)})^2} dx$$
I know that tan(x)=t works, but there are some problems with factorizing denominator and even after factorizing that integral isn't very simple. Does anyone see here any simplier substitution? thank you

2. Jun 21, 2005

### maverick280857

Its $tan(\frac{x}{2}) = t$. Try it. What are the problems?

3. Jun 21, 2005

### dextercioby

Don't be afraid of this simple algebra.When doing integrals,you ought to be prepared for worst.

BTW,LaTex acknowledges the function "tangent",as long as you write it " \tan ".

Daniel.

4. Jun 21, 2005

### Yegor

In this case denominator is $$t^4-t^3+2t^2+t+1$$. This i can't factorize also. Even Mathematica doesn't help.

5. Jun 21, 2005

### Yegor

Hm. Yes, i'm probably just too lazy today to work on algebra. But i have one more question about such problems. If i have polynomials with complex solutions what is the algorithm for factorizing them?
As examples:1) $$t^4-t^3+2t^2+t+1$$
2) $$4t^4+4t^3+9t^2+4t+4$$

6. Jun 21, 2005

### dextercioby

It can't be.

$$I= \int \frac{dx}{\left(\sin x+\frac{2}{\cos x}\right)^{2}}$$ (1)

Make the sub

$$\tan\frac{x}{2}=t$$ (2)

An you'll get the integral

$$I=2\int \frac{\cos ^2\left( 2\arctan t\right) }{\left[ \sin \left( 2\arctan t\right) \cos \left( 2\arctan t\right) +2\right] ^2\left( 1+t^2\right) }\,dt$$ (3)

Then

$$\left\{\begin{array}{c} \cos \left( \arctan t\right) =\frac{1}{\sqrt{\left( 1+t^2\right) }}\\ \sin \left( \arctan t\right) =\frac {t}{\sqrt{\left( 1+t^2\right) }} \end{array} \right$$ (4)

Can you take it from here...?

Daniel.

Last edited: Jun 21, 2005
7. Jun 21, 2005

### maverick280857

Yeah I forgot. Thanks anyway.

8. Jun 21, 2005

### Yegor

$$\left{\begin{array}{c} \cos \left( 2\arctan t\right) =\frac{1-t^2}{{\left( 1+t^2\right) }}\\ \sin \left( 2\arctan t\right) =\frac {2t}{{\left( 1+t^2\right) }} \end{array} \right$$

9. Jun 21, 2005

### Yegor

Yes, i did it dextercioby. This i have from it:
$$\int\frac{(1+t^2)(1-t^2)^2}{t^4-t^3+2t^2+t+1}$$
OK. i can devide this fraction, but i can do nothing with denominator

10. Jun 21, 2005

### dextercioby

Here's what i'd do.I denote your integral by "I".

$$I=:\int\frac{\cos^{2}x}{\left(\frac{1}{2}\sin 2x +2\right)^{2}} \ dx$$ (1)

I define its sister integral

$$J=:\int\frac{\sin^{2}x}{\left(\frac{1}{2}\sin 2x +2\right)^{2}} \ dx$$ (2)

The difference

$$I-J=\int \frac{\cos 2x}{\left(\frac{1}{2}\sin 2x +2\right)^{2}} \ dx =\int \frac{d\left(\frac{1}{2}\sin 2x +2\right)}{\left(\frac{1}{2}\sin 2x +2\right)^{2}} =-\frac{1}{\frac{1}{2}\sin 2x +2}$$ (3)

I'll concentrate on the sum.

To be continued.

Daniel.

11. Jun 21, 2005

### dextercioby

Here's the sum

$$I+J=\int \frac{dx}{\left(\frac{1}{2}\sin 2t +2\right)^2}$$ (4)

I make the sub

$$2x=t$$ (5)

$$I+J=\frac{1}{2}\int \frac{dt}{\left(\frac{1}{2}\sin t +2\right)^{2}}$$ (6)

I make the sub

$$\tan\frac{t}{2}=u$$ (7)

$$I+J=\int \frac{1+u^{2}}{(2u^{2}+u+2)^{2}} \ du=\frac{1}{4}\int \frac{1+u^{2}}{\left[\left(u+\frac{1}{4}-\frac{i\sqrt{3}}{2}\right)\left(u+\frac{1}{4}+\frac{i\sqrt{3}}{2}\right)\right]^{2}} \ du$$ (8)

Therefore

$$I+J=\int \frac{1+u^{2}}{\left[u^{2}+u\left(\frac{1}{2}-i\sqrt{3}\right)-\frac{11}{16}-\frac{i\sqrt{3}}{4}\right]\left[u^{2}+u\left(\frac{1}{2}+i\sqrt{3}\right)-\frac{11}{16}+\frac{i\sqrt{3}}{4}\right]} \ du$$ (9)

and now apply simple fractions.

Daniel.