1. ## more integration help please

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 $\displaystyle (2- 2sin(x))^{1/2}$ you will have $\displaystyle \int\frac{2cos(x)dx}{(2- 2sin(x))^{1/2}}= $$\displaystyle \sqrt{2}\int\frac{cos(x)dx}{(1- sin(x))^{1/2}}} and the substitution \displaystyle u= 1- sin(x) looks like something to try. 3. ## Re: more integration help please Originally Posted by HallsofIvy I don't know what you mean by "partial integration". integration by "partial fractions"? Have you tried doing what it suggests? Once you multiply both numerator and denominator by \displaystyle (2- 2sin(x))^{1/2} you will have \displaystyle \int\frac{2cos(x)dx}{(2- 2sin(x))^{1/2}}=$$\displaystyle \sqrt{2}\int\frac{cos(x)dx}{(1- sin(x))^{1/2}}}$ and the substitution $\displaystyle u= 1- sin(x)$ looks like something to try.
I honestly did try. But I didn't get what you got. But now that I see it I realize my mistake. Sometimes I forget that as long as they're both under the radical sign you can combine terms underneath one radical sign. Thanks for the help!

4. ## Re: more integration help please

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.