Answer
$$\int^{\pi/4}_0\frac{1+\cos^2\theta}{\cos^2\theta}d\theta=1+\frac{\pi}{4}$$
Work Step by Step
$$A=\int^{\pi/4}_0\frac{1+\cos^2\theta}{\cos^2\theta}d\theta$$ $$A=\int^{\pi/4}_0(\frac{1}{\cos^2\theta}+1)d\theta$$ $$A=\int^{\pi/4}_0(\sec^2\theta+1)d\theta$$ According to Table 1, we have $$\int \sec^2 xdx=\tan x+C$$ $$\int kdx=kx+C$$
Therefore, $$A=(\tan\theta+\theta)\Bigg]^{\pi/4}_0$$ $$A=(\tan\frac{\pi}{4}+\frac{\pi}{4})-(\tan0+0)$$ $$A=(1+\frac{\pi}{4})-(0+0)$$ $$A=1+\frac{\pi}{4}$$
