Integration using inverse trig indentities?

Cpt Qwark
Messages
45
Reaction score
1

Homework Statement


1.\int{\frac{sinx}{1+cos^{2}x}} \, dx
2.\int{\frac{1}{13-4x+x^2}} \, dx

Homework Equations


Inverse trig identities.

The Attempt at a Solution


For the first one, I'm not too sure about what to do with the sinx on the numerator and i have tried u-substitution to no avail (different answer). For the second one, I tried factorising but only had solutions for x\inℂ.
 
Physics news on Phys.org
You can maybe try to start with the substitution u = cos(x) and reforming the integral. A second substitution will be necessary after the first substitution.
 
eple said:
You can maybe try to start with the substitution u = cos(x) and reforming the integral. A second substitution will be necessary after the first substitution.
Second substitution is not necessary. It can be done in that first one.
 
For the second integral start by completing the square in the denominator.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top