# What is the relation in terms of the integral $\int_{b}^{a}(cos( b)-cos( x))^\frac{1}{2}dx$?

I am considering the relation between the integral$$I_1=\int_{b}^{a}(cos( b)-cos( u))^\frac{1}{2}du$$ and $$I_2=\int_{b}^{a}(cos(k\cdot b)-cos(k\cdot v))^\frac{1}{2}dv$$where $a>b>0$ and $k>0$ ?

I tried to find the antiderivative but could not be found. I also changed the variable by $u=k\cdot v$, then I got $I_1=k\int_{\frac{b}{k}}^{\frac{a}{k}}(cos( b)-cos( k\cdot v))^\frac{1}{2}dv$. However, I didn't see any relation with $I_2.$

I wondering what are the relations between them. Can I use one to represent the other one?

Any help would be appreciated! Thanks!

• As long as the bounds are the same and that you face elliptic integrals ... – Claude Leibovici Apr 10 '18 at 3:19

The sum to product identity yields $$\cos(kb)-\cos(kv) = -2\sin\bigg(k\frac{b+v}{2}\bigg)\sin\bigg(k\frac{b-v}{2}\bigg)$$ this might help.
Even the antiderivatives are not related because of the $k$ factor in the constant term $$J_1=\int\sqrt{\cos( b)-\cos( u)}\,du$$ $$J_2=\int\sqrt{\cos(k b)-\cos(k v)}\,dv=\frac{1}{k}\int\sqrt{\cos (b k)-\cos (u)}\,du$$ Moreover, these are very difficult antiderivatives since they invoke elliptic integrals of the second kind (on purpose, I did not simplify the result) $$K=\int\sqrt{a-\cos( x)}\,dx=2\frac{ \sqrt{a-\cos (x)} }{\sqrt{\frac{a-\cos (x)}{a-1}}}E\left(\frac{x}{2}|-\frac{2}{a-1}\right)$$ (take care about the notations - these are those used by Wolfram Alpha and Mathematica)