# Integral Help

1. Mar 26, 2008

### noblerare

I've been having trouble with the following three integrals. If anybody could give me a hint on how to get started, that would be greatly appreciated.

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

$$\int$$$$\frac{dx}{1+cos^2(x)}$$

2. The attempt at a solution

I've tried writing it in these ways:

$$\int\frac{dx}{2-sin^2(x)}$$

$$\int\frac{dx}{(secx+cosx)(cosx)}$$

Let u=cosx
$$\int\frac{-du}{\sqrt{1-u^2}(1+u^2)}$$

So far I've gotten nowhere. I've also tried to make the integral look like the integral for arctanx. I realize that I need to have a -sinx in the numerator but don't know how to make it work out.

Problem 2:
1. The problem statement, all variables and given/known data

$$\int$$$$\frac{dx}{sec^2(x)+tan^2(x)}$$

2. The attempt at a solution

I've tried writing it like:

$$\int$$$$\frac{cos^2(x)}{1+sin^2(x)}$$

U-substitution and trig-substition don't seem to work.

Problem 3:
1. The problem statement, all variables and given/known data

$$\int$$$$\frac{dx}{x^3+1}$$

2. The attempt at a solution

In this problem I actually got somewhere:

I split the denominator into two polynomials and used partial fractions.

I ended with:

$$\frac{1}{3}$$ln(x+1) - $$\frac{1}{3}$$$$\int$$$$\frac{(x-2)dx}{x^2-x+1}$$

Then:

$$\frac{1}{3}$$ln(x+1) - $$\frac{1}{6}$$$$\int$$$$\frac{(2x-4)dx}{x^2-x+1}$$

$$\frac{1}{3}$$ln(x+1) - $$\frac{1}{6}$$$$\int$$$$\frac{(2x-1)dx}{x^2-x+1}$$ - $$\int$$$$\frac{-3dx}{x^2-x+1}$$

$$\frac{1}{3}$$ln(x+1) - $$\frac{1}{6}$$ln(x^2-x+1)+3$$\int$$$$\frac{dx}{x^2-x+1}$$

Now I'm stuck.... don't know what to do.

Any help would be greatly appreciated!

2. Mar 26, 2008

### Gib Z

For the first two, trying $$t= \tan (x/2)$$ should work nicely =]

Third one, for the final integral, complete the square in the bottom and get it into the standard arctan form.

3. Mar 27, 2008

### noblerare

Hey thanks for the help.

For the integral:

$$\int$$$$\frac{dx}{1+cos^2(x)}$$

I use u=tan(x/2) so, after much simplification, I get...

$$\int$$$$\frac{2(1+u^2)du}{(1+u^2)^2+(1-u^2)^2}$$

$$\int$$$$\frac{(1+u^2)du}{(1+u^4)}$$

Now I don't know exactly what I should do. If I try trig substituion, I end up with..

$$\frac{1}{2}$$$$\int$$$$\frac{d\theta}{\sqrt{tan\theta}}$$ + $$\frac{1}{2}$$$$\int$$$$\frac{tan\thetad\theta}{2\sqrt{tan\theta}}$$

Now I don't really know what I should do from here...

4. Mar 28, 2008

### Gib Z

You should check your simplification, i get $$\int \frac{ 2(u^2+1)}{1+ (u^2-1)^2} } du$$ which can be done with some more partial fractions.

5. Mar 28, 2008

### tiny-tim

Hi noblerare!

Yes, Gib Z's tan(x/2) method is a good one, which you should definitely remember, and it certainly works in this case.

But in this case, the square means that tanx might work even better (or cotx).

Alternatively:

Hint: mutiply both top and bottom by cosec^2(x) … now what does that remind you of … ?

6. Mar 28, 2008

### noblerare

Yay! I finally got it. Thank you so much, Gib Z and tiny-tim!