how would you integrate x.secx.tanx dx??
thanks for any help
$\displaystyle \int x\sec{x}\tan{x}~dx = \int x(\sec{x})'~dx = x\sec{x} - \int \sec{x}~dx$
It can be easily shown that $\displaystyle \int \sec{x}~dx = \ln|\sec{x}+\tan{x}|$ by multiplying the integral expression with $\displaystyle \frac{\sec{x}+\tan{x}}{\sec{x}+\tan{x}}$