I = \Bigl.x\cdot\Bigl(-\tfrac{1}{4}\cos(4x)\Bigr)\Bigr|_{0}^{\frac{\pi}{4}} \;-\;\int_{0}^{\frac{\pi}{4}}\Bigl(-\tfrac{1}{4}\cos(4x)\Bigr)\,dx.
Isto é
I = -\frac{x}{4}\cos(4x)\Bigl|_{0}^{\frac{\pi}{4}} +\frac{1}{4}\int_{0}^{\frac{\pi}{4}}\cos(4x)\,dx.
Passo 4
Agora calculamos \displaystyle \int \cos(4x)\,dx = \frac{1}{4}\operatorname{sen}(4x). Logo
I = -\frac{x}{4}\cos(4x)\Bigl|_{0}^{\frac{\pi}{4}} +\frac{1}{4}\cdot\frac{1}{4}\operatorname{sen}(4x)\Bigl|_{0}^{\frac{\pi}{4}} = -\frac{x}{4}\cos(4x)\Bigl|_{0}^{\frac{\pi}{4}} +\frac{1}{16}\operatorname{sen}(4x)\Bigl|_{0}^{\frac{\pi}{4}}.
Passo 5
Avaliaremos em x=\frac{\pi}{4} e em x=0:
Para x=\frac{\pi}{4}: \cos(4\cdot\frac{\pi}{4})=\cos(\pi)=-1 e \operatorname{sen}(\pi)=0.
Então -\tfrac{\pi/4}{4}\cdot(-1)+\tfrac{1}{16}\cdot0 = \tfrac{\pi}{16}.
Para x=0: \cos(0)=1 e \operatorname{sen}(0)=0.
Então -\tfrac{0}{4}\cdot1+\tfrac{1}{16}\cdot0 = 0.