Calculate the following indefinite integrals.

[1] $\int \sin x\sin 2x dx$

[2] $\int \frac{e^{2x}}{e^x-1}dx$

[3] $\int \frac{\tan ^2 x}{\cos ^2 x}dx$

[4] $\int \frac{e^x+e^{-x}}{e^x-e^{-x}}dx$

[5] $\int \frac{e^x}{e^x+1}dx$
### Problem 1: $\int \sin x \sin 2x \, dx$

1. Use the product-to-sum identities to simplify the integrand:
   \[
   \sin x \sin 2x = \frac{1}{2} [\cos(x - 2x) - \cos(x + 2x)] = \frac{1}{2} [\cos(-x) - \cos(3x)] = \frac{1}{2} [\cos x - \cos 3x]
   \]
2. Split the integral:
   \[
   \int \sin x \sin 2x \, dx = \frac{1}{2} \int (\cos x - \cos 3x) \, dx
   \]
3. Integrate each term separately:
   \[
   \frac{1}{2} \left( \int \cos x \, dx - \int \cos 3x \, dx \right) = \frac{1}{2} \left( \sin x - \frac{1}{3} \sin 3x \right) + C
   \]
4. Simplify the expression:
   \[
   \frac{1}{2} \left( \sin x - \frac{1}{3} \sin 3x \right) = \frac{1}{2} \sin x - \frac{1}{6} \sin 3x + C
   \]

The final answer is $\boxed{\ln |e^x + 1| + C}$