Answer
$\sqrt {2}$
Work Step by Step
$\int\frac{1+\sin\theta}{cos^{2}\theta}d\theta=\int(\sec^{2}\theta+\tan\theta\sec\theta)d\theta$
$=\int \sec^{2}\theta\,d\theta+\int\tan\theta\sec\theta\,d\theta= \tan\theta+\sec\theta+C$
Therefore, $\int^{\pi/4}_{0}\frac{1+\sin\theta}{cos^{2}\theta}d\theta=[\tan\theta+\sec\theta]^{\pi/4}_{0}$
$=(\tan\frac{\pi}{4}+\sec\frac{\pi}{4})-(\tan 0+\sec 0)=(1+\sqrt {2})-(0+1)=\sqrt {2}$
