# Integration by trigonometric substitution

## Homework Statement

∫ 1/(x√(25-x^2)) dx

## Homework 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)

## 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?

## The Attempt at a Solution

Related Calculus and Beyond Homework Help News on Phys.org
LCKurtz
Homework Helper
Gold Member
Differentiate the answers and see if you get the integrand. That will tell you.

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