# Integration by trigonometric substitution

1. Oct 17, 2011

### cesaruelas

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

∫ 1/(x√(25-x^2)) dx

2. Relevant equations

by trigonometic substitution I got the following equations
x = 5 sin θ
dx = 5 cos θ dθ
√(25-x^2) = 5 cos θ
θ = sin^-1 (x/5)

3. The attempt at a solution

I solved it and got the following (so far, it is okay with wolfram alpha):

∫ 1/(x√(25-x^2)) dx = (1/5) ∫ csc θ dθ

according to me, the result should be (1/5) ( ln ( 5/x - √(25-x^2)/x ) )
but according to wolfram alpha it should be (1/5) ( ln (x) - ln ( √(25-x^2) + 5 ) )

are these two results equivalent?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 17, 2011

### LCKurtz

Differentiate the answers and see if you get the integrand. That will tell you.

3. Oct 17, 2011

### SammyS

Staff Emeritus
$\displaystyle \ln (x) - \ln ( \sqrt{25-x^2} + 5 )=\ln\left(\frac{x}{\sqrt{25-x^2} + 5}\right)$
$\displaystyle =-\ln\left(\frac{\sqrt{25-x^2}}{x}+\frac{5}{x}\right)$​

So, there's only a disagreement about the sign.