d1di9u01

2023-02-25

How to find the vertex of this parabola $y=-{x}^{2}-2x-5$?

Julien Zavala

Beginner2023-02-26Added 7 answers

Find the integral of ${\mathrm{tan}}^{2}x$.

$\int {\mathrm{tan}}^{2}xdx=\int \left(se{c}^{2}x-1\right)dx\left(\because se{c}^{2}x-ta{n}^{2}x=1\right)\int {\mathrm{tan}}^{2}xdx=\mathrm{tan}x-x+C\left(\because \int se{c}^{2}xdx=\mathrm{tan}x,\int 1dx=x.\right)$

Thus, the required integral is $\int {\mathrm{tan}}^{2}xdx=\mathrm{tan}x-x+C$ ,where $C$ is a constant of integration .

$\int {\mathrm{tan}}^{2}xdx=\int \left(se{c}^{2}x-1\right)dx\left(\because se{c}^{2}x-ta{n}^{2}x=1\right)\int {\mathrm{tan}}^{2}xdx=\mathrm{tan}x-x+C\left(\because \int se{c}^{2}xdx=\mathrm{tan}x,\int 1dx=x.\right)$

Thus, the required integral is $\int {\mathrm{tan}}^{2}xdx=\mathrm{tan}x-x+C$ ,where $C$ is a constant of integration .