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Question Number 165320 by mnjuly1970 last updated on 29/Jan/22

$$$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{4}}cos\left(2x\right).e^{\lfloor sin\left(x\right)+cos\left(x\right)\rfloor }dx$$$$\lfloor x\rfloor =max\{m\in \mathbb{Z}\mid m⩽x\}$$$$----$$

Answered by mahdipoor last updated on 30/Jan/22

$$sinx+cosx=\sqrt{2}sin(\frac{\pi }{4}+x)=A$$$$0⩽x⩽\frac{\pi }{4}\Rightarrow 1⩽A⩽\sqrt{2}\Rightarrow \lfloor A\rfloor =1$$$$\mathrm{\Omega }={\int }_0^{\frac{\pi }{4}}cos\left(2x\right).e^1.dx=e{\left[\frac{sin\left(2x\right)}{2}\right]}_0^{\pi /4}=\frac{e}{2}$$

Commented by mnjuly1970 last updated on 30/Jan/22

$$mamnoonjenabemahdipoor$$$$verynicesolution$$

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