Answer
$$\int^{\pi/4}_0(\sec \theta\tan \theta)d\theta=\sqrt 2-1$$
Work Step by Step
$$A=\int^{\pi/4}_0(\sec \theta\tan \theta)d\theta$$ According to Table 1, we have $$\int \sec x\tan xdx=\sec x+C$$
Therefore, $$A=(\sec\theta)\Bigg]^{\pi/4}_0$$ $$A=(\sec\frac{\pi}{4}-\sec0)$$ $$A=\sqrt 2-1$$