∫0sec2π4θ dθ\int\limits_{0\sec^2}^{\frac{\pi}{4}} \theta \, d\theta0sec2∫4πθdθ
= [tanθ]0π4\left[\tan\theta\right]_{0}^{\frac{\pi}{4}}[tanθ]04π
= tanπ4−tan0\tan\frac{\pi}{4} - \tan 0tan4π−tan0
= 1−01 - 01−0
= 1.
