burkinaval1b

2021-12-17

What's the integral of $\int \mathrm{sec}x\mathrm{tan}xdx$?

Medicim6

Expert

$\int \mathrm{sec}\left(x\right)\mathrm{tan}\left(x\right)dx=\mathrm{sec}\left(x\right)+C$
Here is the explanation:
$\int \mathrm{sec}\left(x\right)\mathrm{tan}\left(x\right)dx=\int \frac{1}{\mathrm{cos}\left(x\right)}×\frac{\mathrm{sin}\left(x\right)}{\mathrm{cos}\left(x\right)}dx=\int \frac{\mathrm{sin}\left(x\right)}{{\left(\mathrm{cos}\left(x\right)\right)}^{2}}dx=\frac{1}{\mathrm{cos}\left(x\right)}+C=\mathrm{sec}\left(x\right)+C$

Rita Miller

Expert

$u=\mathrm{sec}\left(x\right)$
$du={\left(\frac{1}{\mathrm{cos}\left(x\right)}\right)}^{\prime }=-\frac{1}{{\mathrm{cos}}^{2}\left(x\right)}×\left(-\mathrm{sin}\left(x\right)dx\right)=\frac{\mathrm{sin}\left(x\right)}{{\mathrm{cos}}^{2}\left(x\right)}dx=\mathrm{tan}\left(x\right)\mathrm{sec}\left(x\right)dx$
Thus,
$\int \mathrm{tan}\left(x\right)\mathrm{sec}\left(x\right)dx=\int du=u=\mathrm{sec}\left(x\right)+C$

nick1337

Expert

$\int \mathrm{sec}\left(x\right)\mathrm{tan}\left(x\right)=\int d\left(\mathrm{sec}\left(x\right)\right)=\mathrm{sec}\left(x\right)+C$