Need help with the integral $\int \frac{2\tan(x)+3}{5\sin^2(x)+4}\,dx$ I'm having a problem resolving the following integral, spent almost all day trying.
Any help would be appreciated.
$$\int \frac{2\tan(x)+3}{5\sin^2(x)+4}\,dx$$

Hint:


*

*Multiply the numerator and denomiator by $\sec^{2}(x)$. Then you have $$\int \frac{2 \tan{x} +3}{5\sin^{2}(x)+4} \times \frac{\sec^{2}(x)}{\sec^{2}{x}} \ dx$$

*Now, the denomiator becomes $5\tan^{2}(x) + 4\cdot \bigl(1+\tan^{2}(x)\bigr)$, and  you can put $t=\tan{x}$.

*So your new integral in terms of $t$ is : $\displaystyle \int \frac{2t+3}{9t^{2}+4} \ dt = \int\frac{2t}{9t^{2}+4} \ dt + \int\frac{3}{9t^{2}+4} \ dt$.
