How to calculate this elliptic integral? I need to calculate this integral
$\int \cos(t\theta)\sqrt{1-k^{2}cos(\theta)^{2}} \mathrm{d}\theta$
But according to Wolfram|Alpha there is no result in terms of standart mathematical functions.
How do we calculate this integral?
Edit: t is integer.
 A: $$I_n=\int_0^\varphi\cos n\theta\sqrt{1-k^2\cos^2\theta}\,d\theta=\int_0^\varphi T_n(\cos\theta)\sqrt{1-k^2\cos^2\theta}\,d\theta$$
where $T_n$ is a Chebyshev polynomial of the first kind. If $n$ is odd, $I_n$ is elementary because then $T_n(\cos\theta)=P_n(\sin^2\theta)\cos\theta$ with $P_n$ a polynomial; substituting $u=\sin\theta$ gives
$$I_n=\int_0^{\sin\varphi}P_n(u^2)\sqrt{1-k^2(1-u^2)}\,du$$
which is easily integrated. If $n$ is even, however, $I_n$ requires elliptic integrals: express $T_n(\cos\theta)=P_n(\sin^2\theta)$, then letting $\nu^2=\frac{k^2}{1-k^2}\ge0$,
$$I_n=\sqrt{1-k^2}\int_0^\varphi P_n(\sin^2\theta)\sqrt{1+\nu^2\sin^2\theta}\,d\theta=\int_0^\varphi\frac{P_n(\sin^2\theta)(1+\nu^2\sin^2\theta)}{\sqrt{1+\nu^2\sin^2\theta}}\,d\theta$$
This is a linear combination of integrals of the form
$$\int_0^\varphi\frac{\sin^{2j}\theta}{\sqrt{1+\nu^2\sin^2\theta}}\,d\theta=(1+\nu^2)^{-j-1/2}\int_0^{u_1}\operatorname{sd}^{2j}u\,du=(1-k^2)^{j+1/2}M_{2j}$$
The middle relation is Byrd and Friedman 282.04. The hidden parameter of the Jacobian elliptic function $\operatorname{sd}$ is $m=\frac{\nu^2}{1+\nu^2}=k^2$ and
$$\operatorname{sn}u_1=\sin\psi=\sqrt{\frac{(1+\nu^2)\sin^2\varphi}{1+\nu^2\sin^2\varphi}}$$
The integrals $M_{2j}$ are given by B&F 318.00, .02, .05:
$$M_0=F(\psi,m)\qquad M_2=\frac{E(\psi,m)-(1-m)u_1-m\operatorname{sn}u_1\operatorname{cd}u_1}{m(1-m)}$$
$$M_{2j+2}=\frac{2j(2m-1)M_{2j}+(2j-1)M_{2j-2}-\operatorname{cd}u_1\operatorname{nd}u_1\operatorname{sd}^{2j-1}u_1}{(2j+1)m(1-m)}$$
