Question Number 159491 by akolade last updated on 17/Nov/21
Answered by Tokugami last updated on 18/Nov/21
![∫_0 ^( (d/dx)∫_0 ^(lim_(x→∞) e^(−x) Σ_(k=0) ^∞ (x^k /(k!))) π dt) (dr/(1+8sin^2 (tan(x))))=θ Σ_(k=0) ^∞ (x^k /(k!))=e^x e^(−x) e^x =1 lim_(x→∞) 1=1 ∫_0 ^( 1) π dt=πt]_0 ^1 =π−0=π (d/dx)(π)=0 ∫_0 ^( 0) (dr/(1+8sin^2 (tan(r))))=0 θ=0 (((cos θ sin θ)^2 )/2)=((((1)(0))^2 )/2) =0](https://www.tinkutara.com/question/Q159505.png)
$$\int_{\mathrm{0}} ^{\:\frac{{d}}{{dx}}\int_{\mathrm{0}} ^{\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{e}^{−{x}} \underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{k}} }{{k}!}} \pi\:{dt}} \frac{{dr}}{\mathrm{1}+\mathrm{8sin}^{\mathrm{2}} \left(\mathrm{tan}\left({x}\right)\right)}=\theta \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{x}^{{k}} }{{k}!}={e}^{{x}} \\ $$$${e}^{−{x}} {e}^{{x}} =\mathrm{1} \\ $$$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\mathrm{1}=\mathrm{1} \\ $$$$\left.\int_{\mathrm{0}} ^{\:\mathrm{1}} \pi\:{dt}=\pi{t}\right]_{\mathrm{0}} ^{\mathrm{1}} =\pi−\mathrm{0}=\pi \\ $$$$\frac{{d}}{{dx}}\left(\pi\right)=\mathrm{0} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{0}} \frac{{dr}}{\mathrm{1}+\mathrm{8sin}^{\mathrm{2}} \left(\mathrm{tan}\left({r}\right)\right)}=\mathrm{0} \\ $$$$\theta=\mathrm{0} \\ $$$$\frac{\left(\mathrm{cos}\:\theta\:\mathrm{sin}\:\theta\right)^{\mathrm{2}} }{\mathrm{2}}=\frac{\left(\left(\mathrm{1}\right)\left(\mathrm{0}\right)\right)^{\mathrm{2}} }{\mathrm{2}} \\ $$$$=\mathrm{0} \\ $$