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Question Number 159491 by akolade last updated on 17/Nov/21

Answered by Tokugami last updated on 18/Nov/21

$${\int }_0^{\frac{d}{dx}{\int }_0^{\underset{x\to \mathrm{\infty }}{\lim }{e}^{-x}\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{x^k}{k!}}\pi dt}\frac{dr}{1+{8\sin }^2\left(\tan \left(x\right)\right)}=\theta$$$$\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{x^k}{k!}=e^x$$$$e^{-x}{e}^x=1$$$$\underset{x\to \mathrm{\infty }}{\lim }1=1$$$${{\int }_0^1\pi dt=\pi t]}_0^1=\pi -0=\pi$$$$\frac{d}{dx}\left(\pi \right)=0$$$${\int }_0^0\frac{dr}{1+{8\sin }^2\left(\tan \left(r\right)\right)}=0$$$$\theta =0$$$$\frac{{\left(\cos \theta \sin \theta \right)}^2}{2}=\frac{{\left(\left(1\right)\left(0\right)\right)}^2}{2}$$$$=0$$

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