How Do I Solve These Challenging Integrals?

[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.

Homework Statement

\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:

Homework Statement

\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:

Homework Statement

\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!

 

[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.

 

[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...

 

[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.

 

[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 … ?

 

[noblerare]

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