Question
What is indefinite integral of 1/(x^2-4)^(1/2)?
Answers
Steve
this is a classic case for trig substitution
Let x = 2secu
Then x^2 - 4 = 4sec^2 u - 4 = 4tan^2 u
dx = 2 secu tanu du
1/(x^2-4)^1/2 dx
= 1/(2tan u) * 2 secu tanu du
= secu du
Now, Int(secu du)
= ln(secu + tanu)
= ln(x + sqrt(x^2 - 4)) + C
question: why can it be this, instead of ln((1/2)(x + sqrt(x^2 - 4))) + C ?
Let x = 2secu
Then x^2 - 4 = 4sec^2 u - 4 = 4tan^2 u
dx = 2 secu tanu du
1/(x^2-4)^1/2 dx
= 1/(2tan u) * 2 secu tanu du
= secu du
Now, Int(secu du)
= ln(secu + tanu)
= ln(x + sqrt(x^2 - 4)) + C
question: why can it be this, instead of ln((1/2)(x + sqrt(x^2 - 4))) + C ?