I feel like I should know how to do this but I'm kind of getting stuck

Question: $\displaystyle \int^{\pi/4}_0 \frac{e^{tanx}}{1-sin^2 x}$

So i used the trig identity and got

$\displaystyle \int^{\pi/4}_0 \frac{e^{tanx}}{cos^2 x}$

$\displaystyle let \:u=tanx \:\:\: du=\frac{1}{cosx}$

$\displaystyle \int^{\pi/4}_0 \frac{e^u}{cosx\, du}$

so basically I am left with a spare 1/cosx that I'm not sure what to do with, any help would be appreciated