Integrating ∫√(2+2sinθ) using (2+2sinΘ)(2-sinΘ)

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Homework Statement



How to integrate∫√(2+2sinθ)

Homework Equations



Making use of (2+2sinΘ)(2-sinΘ)=4-4sinΘ^2

The Attempt at a Solution



Multiply and dividing the integrand by √2-2sinΘ
 
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aerograce said:

Homework Statement



How to integrate∫√(2+2sinθ)

Homework Equations



Making use of (2+2sinΘ)(2-sinΘ)=4-4sinΘ^2

The Attempt at a Solution



Multiply and dividing the integrand by √2-2sinΘ

Show us what happened when you did that...
 
LCKurtz said:
Show us what happened when you did that...

It becomes,

∫ 4cosΘ^2/√(2-2sinΘ)
 
aerograce said:
It becomes,

∫ 4cosΘ^2/√(2-2sinΘ)

Isn't there a square root missing in the numerator? Have you tried any substitutions?
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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