How in the world do you solve this???:
int[sec(1-x)*tan(1-x)*dx]
Thank you!
Remember that $\displaystyle \sec{(1 - x)} = \frac{1}{\cos{(1 - x)}}$ and that $\displaystyle \tan{(1 - x)} = \frac{\sin{(1-x)}}{\cos{(1 -x)}}$.
Using these, we see that
$\displaystyle \int{\sec{(1-x)}\,\tan{(1-x)}\,dx} = \int{\frac{1}{\cos^2{(1-x)}}\,\sin{(1-x)}\,dx}$.
Use the substitution $\displaystyle u = \cos{(1-x)}$ so that $\displaystyle \frac{du}{dx} = \sin{(1-x)}$. Then the integral becomes
$\displaystyle \int{\frac{1}{u^2}\,\frac{du}{dx}\,dx}$
$\displaystyle = \int{u^{-2}\,du}$
$\displaystyle = -u^{-1} + C$
$\displaystyle = -\frac{1}{\cos{(1-x)}} + C$
$\displaystyle = -\sec{(1-x)} + C$.