For expressing $\int_0^\alpha\sqrt{a^2\sin^2t+b^2\cos^2t}~dt$ in series expression:
$\int_0^\alpha\sqrt{a^2\sin^2t+b^2\cos^2t}~dt$
$=\int_0^\alpha\sqrt{a^2(1-\cos^2t)+b^2\cos^2t}~dt$
$=\int_0^\alpha\sqrt{a^2+(b^2-a^2)\cos^2t}~dt$
$=\int_0^\alpha|a|\sqrt{1+\dfrac{b^2-a^2}{a^2}\cos^2t}~dt$
For the binomial series of $\sqrt{1+x}$ , e.g. according to http://en.wikipedia.org/wiki/Square_root#Properties, $\sqrt{1+x}=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!x^n}{4^n(n!)^2(1-2n)}$
$\therefore\int_0^\alpha|a|\sqrt{1+\dfrac{b^2-a^2}{a^2}\cos^2t}~dt$
$=\int_0^\alpha\sum\limits_{n=0}^\infty\dfrac{(-1)^n(2n)!|a|(b^2-a^2)^n\cos^{2n}t}{4^n(n!)^2(1-2n)a^{2n}}dt$
$=\int_0^\alpha\left(|a|+\sum\limits_{n=1}^\infty\dfrac{(-1)^n(2n)!|a|(b^2-a^2)^n\cos^{2n}t}{4^n(n!)^2(1-2n)a^{2n}}\right)dt$
Now for $\int\cos^{2n}t~dt$ , where $n$ is any natural number,
$\int\cos^{2n}t~dt=\dfrac{(2n)!t}{4^n(n!)^2}+\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin t~\cos^{2k-1}t}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts, e.g. as shown as http://hk.knowledge.yahoo.com/question/question?qid=7012022000808
$\therefore\int_0^\alpha\left(|a|+\sum\limits_{n=1}^\infty\dfrac{(-1)^n(2n)!|a|(b^2-a^2)^n\cos^{2n}t}{4^n(n!)^2(1-2n)a^{2n}}\right)dt$
$=\left[|a|t+\sum\limits_{n=1}^\infty\dfrac{(-1)^n((2n)!)^2|a|(b^2-a^2)^nt}{4^{2n}(n!)^4(1-2n)a^{2n}}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((2n)!)^2((k-1)!)^2|a|(b^2-a^2)^n\sin t~\cos^{2k-1}t}{4^{2n-k+1}(n!)^4(1-2n)(2k-1)!a^{2n}}\right]_0^\alpha$
$=\left[\sum\limits_{n=0}^\infty\dfrac{(-1)^n((2n)!)^2|a|(b^2-a^2)^nt}{4^{2n}(n!)^4(1-2n)a^{2n}}+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((2n)!)^2((k-1)!)^2|a|(b^2-a^2)^n\sin t~\cos^{2k-1}t}{4^{2n-k+1}(n!)^4(1-2n)(2k-1)!a^{2n}}\right]_0^\alpha$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n((2n)!)^2|a|(b^2-a^2)^n\alpha}{4^{2n}(n!)^4(1-2n)a^{2n}}$
$~~~+\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n((2n)!)^2((k-1)!)^2|a|(b^2-a^2)^n\sin\alpha~\cos^{2k-1}\alpha}{4^{2n-k+1}(n!)^4(1-2n)(2k-1)!a^{2n}}$
You can also try another approach that
$\int_0^\alpha\sqrt{a^2\sin^2t+b^2\cos^2t}~dt$
$=\int_0^\alpha\sqrt{a^2\sin^2t+b^2(1-\sin^2t)}~dt$
$=\int_0^\alpha\sqrt{b^2+(a^2-b^2)\sin^2t}~dt$
$=\int_0^\alpha|b|\sqrt{1+\dfrac{a^2-b^2}{b^2}\sin^2t}~dt$
But you should handle $\int\sin^{2n}t~dt$ , where $n$ is any non-negative integer instead