Can we use the substitution t= ln|(2x+3)*(1-x^2)^(1/2)|?
Will that make this simpler?
How do we integrate {[1/(√(1-x^2) )]*[ln|(2x+3)(√(1-x^2))] } ?
Is there any suitable substitution or do we have to do this by parts?
I don't have any idea to proceed on.
3 answers
how about this:
x = sinθ
√(1-x^2) = cosθ
dx = cosθ dθ
so, dθ = dx/cosθ = dx/√(1-x^2)
Then the integrand is
ln((2sinθ+3)cosθ) dθ
Unfortunately, that doesn't integrate any better. We can't use the same trick from before, since we don't have a definite integral. If we did, it might loosen things up a bit.
x = sinθ
√(1-x^2) = cosθ
dx = cosθ dθ
so, dθ = dx/cosθ = dx/√(1-x^2)
Then the integrand is
ln((2sinθ+3)cosθ) dθ
Unfortunately, that doesn't integrate any better. We can't use the same trick from before, since we don't have a definite integral. If we did, it might loosen things up a bit.
What if the limits are 0-1?
2 answers