I don't know what you mean by "partial integration". integration by "partial fractions"? Have you tried doing what it suggests? Once you multiply both numerator and denominator by you will have and the substitution looks like something to try.
how about this one: integrand (2 + 2sinx)^(1/2) dx? The book says "We could evaluate this integral by multiplying and dividing the integrand by (2-2sinx)^(1/2) Do you use partial integration? It's from Stewart's Early Transcendentals fifth edition page 683
I don't know what you mean by "partial integration". integration by "partial fractions"? Have you tried doing what it suggests? Once you multiply both numerator and denominator by you will have and the substitution looks like something to try.
This is actually an arc length problem for the polar equation r=1+sin(x) (x being theta) After some work you get:
L=integral from 0 to 2 times pi (2+2*sin(x))^(1/2) dx......as above it gets reduced to
L=2^(1/2) times integral from 0 to 2 times pi (cos(x)/(1-sin(x))^1/2) dx
But if u=1-sin(x) don't the limits of integration both equal 1?
The book says the answer is L=8 The figure is a cardioid.