how to solve $\int\sqrt{1-16cos^2(x)}\ dx$

Question

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..

Answer 1

Consider:

$$\int \sqrt{1-16 \left(\cos ^2 x\right)} \, dx=-\frac{\sqrt{15} \sqrt{8 \cos(2x) +7} E\left(x\left|\frac{16}{15}\right.\right)}{\sqrt{-8 \cos(2x) -7}} + k$$

Where $E\left(x\left|\frac{16}{15}\right.\right)$ denotes the Elliptic Integral of the second kind.

Answer 2

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*} $$