how to solve $\int\sqrt{1-16cos^2(x)}\ dx$ How can we solve this integral?
$\int\sqrt{1-16cos^2(x)}\ dx$
I tried to make some substitutions, but I failed to solve it.
For example, I took the substitution
$u=4cos(x)$
$dx=-\frac{du}{\sqrt{16-u^2}}$
hence, the integration will be
$\int-\frac{\sqrt{1-u^2}}{\sqrt{16-u^2}}du$
then I tried the substitution
$z=\sqrt{1-u^2}$
$du=-\frac{z}{\sqrt{1-z^2}}dz$
hence, the integration will be
$\int\frac{z^2 dz}{\sqrt{1-z^2}\sqrt{15+z^2}}$
But I do not know how to complete from this step..

You can easily represent the antiderivative using the indefinite elliptic integrals of the second kind (Jacobi-Form) and complex numbers:
$$
\begin{align*}
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{1 - 16 \cdot \left(1 - \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{1 - 16 + 16 \cdot \sin\left( x \right)^{2}} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{-15 + 16 \cdot \sin\left( x \right)^{2}} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{-\left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{(-1) \cdot \left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \sqrt{-1} \cdot \sqrt{\left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \int_{}^{} \mathrm{i} \cdot \sqrt{\left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot \int_{}^{} \sqrt{\left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot \int_{}^{} \sqrt{1 \cdot \left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot \int_{}^{} \sqrt{\frac{15}{15} \cdot \left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot \int_{}^{} \sqrt{15} \cdot \sqrt{\frac{1}{15} \cdot \left( 15 - 16 \cdot \sin\left( x \right)^{2} \right)} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot  \sqrt{15} \cdot \int_{}^{} \sqrt{1 - \frac{16}{15} \cdot \sin\left( x \right)^{2}} \, \operatorname{d}x\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot  \sqrt{15} \cdot \left( \int_{0}^{x} \sqrt{1 - \frac{16}{15} \cdot \sin\left( x \right)^{2}} \, \operatorname{d}x + c \right)\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= \mathrm{i} \cdot  \sqrt{15} \cdot \operatorname{E}\left( x, ~\frac{16}{15} \right) + c\\
\\
\int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x &= c + \sqrt{15} \cdot \operatorname{E}\left( x, ~\frac{16}{15} \right) \cdot \mathrm{i}\\
\Re\left( \int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x \right) &= \Re(c)\\
\Im\left( \int_{}^{} \sqrt{1 - 16 \cdot \cos\left( x \right)^{2}} \, \operatorname{d}x \right) &= \Im(c) + \sqrt{15} \cdot \operatorname{E}\left( x, ~\frac{16}{15} \right)\\
\end{align*}
$$
