L'intégrale simple

Sachant que \((\tan\theta ) ' = 1 + \tan^2\theta ,\) nous modifions \(\tan^3\theta \)en

\(\tan^3\theta = \tan^2\theta \tan\theta = [ ( 1 + \tan^2\theta ) - 1 ] \tan\theta\)

d'où

\(I_4=\int_0^{\pi/4}[(1+\tan^2\theta)-1]\tan\theta d\theta\)

\(I_4=\int_0^{\pi/4}(1+\tan^2\theta)\tan\theta d\theta - \int_0^{\pi/4}\tan\theta d\theta~~\color{red}\text{ (1 pt + 1 pt)}\)

\(I_4=\int_0^{\pi/4}\tan\theta d(\tan\theta) - \int_0^{\pi/4}\frac{-d(\cos\theta)}{\cos\theta}\)

\(=[\frac12\tan^2\theta+\ln|\cos\theta|]_0^{\pi/4}\)

\(I_4=\frac12\tan^2\frac{\pi}4+\ln\cos\frac{\pi}4\)

\(=\color{blue}\frac12+\ln\frac{\sqrt2}2\)

\(=\color{blue}\frac12(1-\ln 2)~~\color{red}\text{ (3 pts)}\)