Solving $\int\frac{\cos^3(2x)}{\sin^2(2x)-1}dx$ I was given the following problem: $$\int\frac{\cos^3(2x)}{\sin^2(2x)-1}\text{dx}$$I found myself stuck and turned to https://www.integral-calculator.com/ for help. It simplified my problem to:$$\int-\cos(2x)\text{dx}$$Obviously, this simplified my problem a great deal and I was then able to solve it. However, I am left with a question - which perhaps is simple to those with a stronger trig background: How is this simplification done? What properties does it use?
 A: In the denominator we have that
$$\sin^2 (2x) -1 = - \big( 1-\sin^2(2x) \big) = -\cos^2(2x)$$
Take the minus sign out of the integral and simplify two factors above with the two below.
A: $\sin^{2} y -1=-(1-\sin^{2} y)=- \cos^{2}y$. This is all you need.
A: I'm going to spell out all the steps explicitly for other people who might find this question.
$$ \int \frac{\cos^3(2x)}{\sin^2(2x) - 1} \mathrm{d}x $$
$$ \int \frac{\cos^3(2x)}{\sin^2(2x) - (\cos^2(2x) + \sin^2(2x))} \mathrm{d}x $$
$$ \int \frac{\cos^3(2x)}{\sin^2(2x) - \sin^2(2x) - \cos^2(2x)} \mathrm{d}x $$
$$ \int \frac{\cos^3(2x)}{- \cos^2(2x)} \mathrm{d}x $$
(-1) is a constant, move it outside the integral.
$$ - \int \frac{\cos^3(2x)}{\cos^2(2x)} \mathrm{d}x $$
simplify. It does not matter that $\cos(2x)$ is sometimes zero since it's only zero at isolated points.
Let's define a new variable $u=2x$ and thus $\mathrm{d}u = 2\mathrm{d}x$ .
$$ -\frac{1}{2} \int \cos(u) \mathrm{d}u $$
$$ -\frac{1}{2} (\sin(u) + C) $$
$$ -\frac{1}{2} \sin(u) + -\frac{1}{2} C $$
Since $C$ is an arbitrary constant of integration, multiplying it by a scalar results in an equivalent expression.
$$ -\frac{1}{2}\sin(u) + C $$
$$ -\frac{1}{2}\sin(2x) + C $$
