- #1
noblerare
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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.
[tex]\int[/tex][tex]\frac{dx}{1+cos^2(x)}[/tex]
2. The attempt at a solution
I've tried writing it in these ways:
[tex]\int\frac{dx}{2-sin^2(x)}[/tex]
[tex]\int\frac{dx}{(secx+cosx)(cosx)}[/tex]
Let u=cosx
[tex]\int\frac{-du}{\sqrt{1-u^2}(1+u^2)}[/tex]
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:
[tex]\int[/tex][tex]\frac{dx}{sec^2(x)+tan^2(x)}[/tex]
2. The attempt at a solution
I've tried writing it like:
[tex]\int[/tex][tex]\frac{cos^2(x)}{1+sin^2(x)}[/tex]
U-substitution and trig-substition don't seem to work.
Problem 3:
[tex]\int[/tex][tex]\frac{dx}{x^3+1}[/tex]
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:
[tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{3}[/tex][tex]\int[/tex][tex]\frac{(x-2)dx}{x^2-x+1}[/tex]
Then:
[tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{6}[/tex][tex]\int[/tex][tex]\frac{(2x-4)dx}{x^2-x+1}[/tex]
[tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{6}[/tex][tex]\int[/tex][tex]\frac{(2x-1)dx}{x^2-x+1}[/tex] - [tex]\int[/tex][tex]\frac{-3dx}{x^2-x+1}[/tex]
[tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{6}[/tex]ln(x^2-x+1)+3[tex]\int[/tex][tex]\frac{dx}{x^2-x+1}[/tex]
Now I'm stuck... don't know what to do.
Any help would be greatly appreciated!
Problem 1.
Homework Statement
[tex]\int[/tex][tex]\frac{dx}{1+cos^2(x)}[/tex]
2. The attempt at a solution
I've tried writing it in these ways:
[tex]\int\frac{dx}{2-sin^2(x)}[/tex]
[tex]\int\frac{dx}{(secx+cosx)(cosx)}[/tex]
Let u=cosx
[tex]\int\frac{-du}{\sqrt{1-u^2}(1+u^2)}[/tex]
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
[tex]\int[/tex][tex]\frac{dx}{sec^2(x)+tan^2(x)}[/tex]
2. The attempt at a solution
I've tried writing it like:
[tex]\int[/tex][tex]\frac{cos^2(x)}{1+sin^2(x)}[/tex]
U-substitution and trig-substition don't seem to work.
Problem 3:
Homework Statement
[tex]\int[/tex][tex]\frac{dx}{x^3+1}[/tex]
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:
[tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{3}[/tex][tex]\int[/tex][tex]\frac{(x-2)dx}{x^2-x+1}[/tex]
Then:
[tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{6}[/tex][tex]\int[/tex][tex]\frac{(2x-4)dx}{x^2-x+1}[/tex]
[tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{6}[/tex][tex]\int[/tex][tex]\frac{(2x-1)dx}{x^2-x+1}[/tex] - [tex]\int[/tex][tex]\frac{-3dx}{x^2-x+1}[/tex]
[tex]\frac{1}{3}[/tex]ln(x+1) - [tex]\frac{1}{6}[/tex]ln(x^2-x+1)+3[tex]\int[/tex][tex]\frac{dx}{x^2-x+1}[/tex]
Now I'm stuck... don't know what to do.
Any help would be greatly appreciated!