Question

∫(dx/1-cos2x)

Answer 1

-(1/2)cot(x) + C

Answer 2

To solve the integral $\int \frac{dx}{1-\cos2x}$, we use trigonometric identities to simplify the integrand.

The key identity is $\cos2x = 1 - 2\sin^2x$. From this, we can express $1-\cos2x$: $1 - \cos2x = 1 - (1 - 2\sin^2x)$ $1 - \cos2x = 1 - 1 + 2\sin^2x$ $1 - \cos2x = 2\sin^2x$

Now, substitute this into the integral: $\int \frac{dx}{1-\cos2x} = \int \frac{dx}{2\sin^2x}$ We can take the constant $\frac{1}{2}$ out of the integral: $= \frac{1}{2} \int \frac{1}{\sin^2x} dx$ We know that $\frac{1}{\sin x} = \csc x$, so $\frac{1}{\sin^2x} = \csc^2x$: $= \frac{1}{2} \int \csc^2x dx$ The standard integral of $\csc^2x$ is $-\cot x$: $= \frac{1}{2} (-\cot x) + C$ $= -\frac{1}{2} \cot x + C$ where $C$ is the constant of integration.